diff --git a/README.md b/README.md index 3fd948f..a02d168 100644 --- a/README.md +++ b/README.md @@ -1,5 +1,3 @@ -![enter image description here](https://drscdn.500px.org/photo/286065929/q=80_m=2000/v2?user_id=20361103&webp=true&sig=5246d3ececc0976199a64920c236729da09957c8ad23afa513084c27681961e4) - | 🍏 | 🍎 | 🍐 | 🍈 | 🥑 | 🥔| 🍠 | 🥝 | 🍱 | 🥞 |🌽| 🥦 | :--------: | :---------: | :---------: | :---------: | :---------: | :---------:| :---------: | :-------: | :-------:| :------:|:------:| :--------: | | [JAVA基础](#JAVA基础) | [JVM知识](#JVM知识)|[开源框架知识](#开源框架知识) | [操作系统知识](#操作系统) |[多线程与并发](#多线程与并发)|[TCP与HTTP](#TCP与HTTP)| [架构设计与分布式](#架构设计与分布式) |[数据结构与算法](#数据结构与算法)|[数据库](#数据库知识)| [消息队列](#消息队列)|[缓存](#缓存) | [搜索](#搜索) @@ -66,13 +64,15 @@ - 简单讲讲tomcat结构，以及其类加载器流程，线程模型等。 - tomcat如何调优，涉及哪些参数 。 + - [Spring中bean的作用域与生命周期](https://mp.weixin.qq.com/s?__biz=MzU4NDQ4MzU5OA==&mid=2247484400&idx=2&sn=7201eb365102fce017f89cb3527fb0bc&chksm=fd985591caefdc872a2fac897288119f94c345e4e12150774f960bf5f816b79e4b9b46be3d7f&token=1990180468&lang=zh_CN#rd) - [讲讲Spring加载流程。](https://segmentfault.com/a/1190000012887776) - [spring循环依赖](http://www.tianxiaobo.com/2018/06/08/Spring-IOC-%E5%AE%B9%E5%99%A8%E6%BA%90%E7%A0%81%E5%88%86%E6%9E%90-%E5%BE%AA%E7%8E%AF%E4%BE%9D%E8%B5%96%E7%9A%84%E8%A7%A3%E5%86%B3%E5%8A%9E%E6%B3%95/) - [Spring AOP的实现原理。](https://www.jianshu.com/p/7fa3e6ebf0c1) - [讲讲Spring事务的传播属性。](https://www.cnblogs.com/cnmenglang/p/6410848.html) - [Spring如何管理事务的。](https://blog.csdn.net/donggua3694857/article/details/69858827) - - 说说你对Spring的理解，非单例注入的原理？它的生命周期？循环注入的原理。 + - 说说你对Spring的理解，非单例注入的原理？循环注入的原理。 - [Springmvc 中DispatcherServlet初始化过程。](https://www.cnblogs.com/xiaoxi/p/6164383.html) + - [SpringMVC 原理了解吗？](https://mp.weixin.qq.com/s?__biz=MzU4NDQ4MzU5OA==&mid=2247484496&idx=1&sn=5472ffa687fe4a05f8900d8ee6726de4&chksm=fd985231caefdb27fc75b44ecf76b6f43e4617e0b01b3c040f8b8fab32e51dfa5118eed1d6ad&token=1990180468&lang=zh_CN#rd) - netty的线程模型，netty如何基于reactor模型上实现的。 - 为什么选择netty。 - [什么是TCP粘包，拆包。解决方式是什么。](https://blog.csdn.net/wxy941011/article/details/80428470) @@ -84,6 +84,7 @@ - [Mybatis的底层实现原理。](https://mp.weixin.qq.com/s/AqqSRYPdUbruSAhhmqVneA) - [69道Spring面试题和答案](http://ifeve.com/spring-interview-questions-and-answers/) - [Spring 面试问题 TOP 50](https://mp.weixin.qq.com/s/eQ-lpYiW7a437djqliKhmg) + - [Spring中用了哪些设计模式](https://mp.weixin.qq.com/s/iGbwKDqB6A0ktNCY8Wd-XQ) ### 操作系统 - Linux系统下你关注过哪些内核参数，说说你知道的。 @@ -92,6 +93,7 @@ - [平时用到哪些Linux命令。以及如何排查线上问题](https://github.com/randian666/algorithm-study/blob/master/MD/Linux.md) - 介绍下你理解的操作系统中线程切换过程。 - 进程和线程的区别。 + - [BATJ面试必会之Linux命令篇](https://mp.weixin.qq.com/s/6ZPNeX9d3avscI642x2kHA) ### 多线程与并发 @@ -100,9 +102,11 @@ - [画一个线程的生命周期状态图。](https://blog.csdn.net/houbin0912/article/details/77969563) - sleep和wait的区别。 - sleep和sleep(0)的区别。 + - [Java 各种锁的小结](https://mp.weixin.qq.com/s/_xazrXa8MBYaz2WaX6BNew) - [Lock与Synchronized的区别 。](https://blog.csdn.net/javazejian/article/details/75043422) - [synchronized的原理是什么](https://blog.csdn.net/javazejian/article/details/72828483) - [量级锁，可重入锁，公平锁，非公平锁，乐观锁，悲观锁。](https://www.toutiao.com/i6630764198357893646/) + - [ReentranLock源码，设计原理，整体过程](https://www.cnblogs.com/chengxiao/p/7255941.html) - 用过哪些原子类，他们的原理是什么。 - JUC下研究过哪些并发工具，讲讲原理。 - [用过线程池吗，如果用过，请说明原理，并说说newCache和newFixed有什么区别，构造函数的各个参数的含义是什么，比如coreSize，maxsize等。](https://blog.csdn.net/wy11933/article/details/80399562) @@ -133,7 +137,7 @@ - [TCP三次握手和四次挥手的流程，为什么断开连接要4次,如果握手只有两次，会出现什么。](https://blog.csdn.net/qzcsu/article/details/72861891) - TIME_WAIT和CLOSE_WAIT的区别。 - [说说你知道的几种HTTP响应码，比如200, 302, 404。](https://blog.csdn.net/ddhsea/article/details/79405996) - - 当你用浏览器打开一个链接（如：http://www.baidu.com）的时候，计算机做了哪些工作步骤。 + - [当你用浏览器打开一个链接的时候，计算机做了哪些工作步骤。](https://segmentfault.com/a/1190000006879700) - [TCP/IP如何保证可靠性，说说TCP头的结构。](https://blog.csdn.net/liuchenxia8/article/details/80428157) - 如何避免浏览器缓存。 - [如何理解HTTP协议的无状态性。](https://blog.csdn.net/tennysonsky/article/details/44562435) @@ -208,7 +212,13 @@ - 一个单向链表，删除倒数第N个数据。 - 200个有序的数组，每个数组里面100个元素，找出top20的元素。 - 单向链表，查找中间的那个元素。 + - [如何判断一个数是否在40亿个整数中？](https://mp.weixin.qq.com/s?__biz=MzU4NDQ4MzU5OA==&mid=2247484394&idx=1&sn=769e86bd815cf836eaa88a46836e4287&chksm=fd98558bcaefdc9d362f4d0912849924ae2d1f1277147edab7508cd3f1753ea4afd719237fa1&scene=21#wechat_redirect) - [剑指Offer学习【所有面试题汇总】](https://blog.csdn.net/derrantcm/article/details/46887821) + - [Leetcode 题解](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Leetcode%20%E9%A2%98%E8%A7%A3.md) + - [算法](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/%E7%AE%97%E6%B3%95.md) + - [Leetcode-LinkList](https://github.com/Snailclimb/JavaGuide/blob/master/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/Leetcode-LinkList1.md) + - [几道常见的子符串算法题](https://github.com/Snailclimb/JavaGuide/blob/master/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/%E6%90%9E%E5%AE%9ABAT%E9%9D%A2%E8%AF%95%E2%80%94%E2%80%94%E5%87%A0%E9%81%93%E5%B8%B8%E8%A7%81%E7%9A%84%E5%AD%90%E7%AC%A6%E4%B8%B2%E7%AE%97%E6%B3%95%E9%A2%98.md) + - [数据结构](https://github.com/Snailclimb/JavaGuide/blob/master/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84.md) ### 数据库 - [MySQL数据库的索引原理、与慢SQL优化的5大原则](https://www.toutiao.com/i6611390023797309960/) - [数据库隔离级别有哪些，各自的含义是什么，MYSQL默认的隔离级别是是什么。](http://www.cnblogs.com/huanongying/p/7021555.html) @@ -230,6 +240,7 @@ - MVCC的含义，如何实现的。 - 你做过的项目里遇到分库分表了吗，怎么做的，有用到中间件么，比如sharding jdbc等,他们的原理知道么。 - MYSQL的主从延迟怎么解决。 + - [MySQL索引背后的数据结构及算法原理](https://www.kancloud.cn/kancloud/theory-of-mysql-index/41846) - [数据库面试问题集锦](https://blog.csdn.net/justloveyou_/article/details/78308460) ### 消息队列 @@ -249,6 +260,7 @@ ### 缓存 - [Redis的高并发和快速原因](https://mp.weixin.qq.com/s/HuwgTvNtmHJ9tpF7V53opg) - [常见的缓存策略有哪些，如何做到缓存(比如redis)与DB里的数据一致性，你们项目中用到了](https://mp.weixin.qq.com/s/yQjTddLWy-oSYPYxyGVeKw) + - [一文揭秘单线程的Redis为什么这么快?](https://mp.weixin.qq.com/s/4ndQI4CyqtjaYJKts8ChdA) - 什么缓存系统，如何设计的。 - [如何防止缓存击穿和雪崩。](https://mp.weixin.qq.com/s/SQQy6NKvoAgIaeAjzkRpDw) - 缓存数据过期后的更新如何设计。 @@ -267,9 +279,12 @@ - 请思考一个方案，设计一个可以控制缓存总体大小的自动适应的本地缓存。 - 如何看待缓存的使用（本地缓存，集中式缓存），简述本地缓存和集中式缓存和优缺点。 - 本地缓存在并发使用时的注意事项。 + - [Redis过期策略及实现原理](https://www.cnblogs.com/linuxprobe-sarah/p/10386800.html) - [redis面试49题](https://mp.weixin.qq.com/s/5mLTVpjn_iebsOp60HFU2Q) ### 搜索 - + - [es 的分布式架构原理能说一下么（es 是如何实现分布式的啊）？](https://doocs.github.io/advanced-java/#/docs/high-concurrency/es-architecture) + - [es 写入数据的工作原理是什么啊？es 查询数据的工作原理是什么啊？底层的 lucene 介绍一下呗？倒排索引了解吗？](https://doocs.github.io/advanced-java/#/docs/high-concurrency/es-write-query-search) + - [es 在数据量很大的情况下（数十亿级别）如何提高查询效率啊？](https://doocs.github.io/advanced-java/#/docs/high-concurrency/es-optimizing-query-performance) - elasticsearch了解多少，说说你们公司es的集群架构，索引数据大小，分片有多少，以及一些调优手段 。 - elasticsearch的倒排索引是什么。 - elasticsearch 索引数据多了怎么办，如何调优，部署。 diff --git a/image/931551357495_.pic.jpg b/image/931551357495_.pic.jpg new file mode 100644 index 0000000..1003422 Binary files /dev/null and b/image/931551357495_.pic.jpg differ diff --git a/image/WechatIMG2430.jpeg b/image/WechatIMG2430.jpeg new file mode 100644 index 0000000..297c61e Binary files /dev/null and b/image/WechatIMG2430.jpeg differ diff --git a/pom.xml b/pom.xml index 472abf1..5496700 100644 --- a/pom.xml +++ b/pom.xml @@ -20,6 +20,46 @@ cglib 2.2.2 + + + + org.testng + testng + 6.14.3 + + + + + com.google.guava + guava + 23.0 + + + + commons-io + commons-io + 2.4 + + + + org.projectlombok + lombok + 1.18.8 + provided + + + + joda-time + joda-time + 2.10.1 + + + + org.apache.commons + commons-lang3 + 3.9 + + diff --git a/src/main/java/com/algorithm/study/demo/LRUCache/LRUCache.java b/src/main/java/com/algorithm/study/demo/LRUCache/LRUCache.java new file mode 100644 index 0000000..0fcbb70 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/LRUCache/LRUCache.java @@ -0,0 +1,152 @@ +package com.algorithm.study.demo.LRUCache; + +import java.util.HashMap; +import java.util.Map; + +/** + * LRU缓存链表实现思路 + * 每次写入数据时将数据放入链表头结点。 + * 使用数据时候将数据移动到头结点。 + * 缓存数量超过阈值时移除链表尾部数据。 + * @Author: liuxun + * @CreateDate: 2018/7/12 下午6:05 + * @Version: 1.0 + */ +public class LRUCache { + class Node{ + private int key; + private int value; + private Node prev; + private Node next; + + public Node(int key,int value){ + this.key=key; + this.value=value; + } + public Node(){} + + public int getKey() { + return key; + } + + public void setKey(int key) { + this.key = key; + } + + public int getValue() { + return value; + } + + public void setValue(int value) { + this.value = value; + } + } + private void moveToHead(Node node){ + remove(node); + addNode(node); + } + //删除尾节点 + private Node popTail(){ + Node prevNode= tail.prev; + tail.prev=prevNode.prev; + prevNode.prev.next=tail; + + prevNode.next=null; + prevNode.prev=null; + + size--; + return prevNode; + } + //删除中间节点 + private void remove(Node node){ + Node prevNode=node.prev; + Node nextNode=node.next; + + prevNode.next=nextNode; + nextNode.prev=prevNode; + + node.next=null; + node.prev=null; + + size--; + } + //添加节点 + private void addNode(Node node){ + node.next=head.next; + node.prev=head; + node.next.prev=node; + head.next=node; + size++; + } + private Map cache=new HashMap(); + private int size=0; + private int capacity=0; + //头结点 + private Node head; + //尾结点 + private Node tail; + public LRUCache(int capacity) { + this.capacity=capacity; + //初始化头尾节点 + this.head=new Node(); + this.tail=new Node(); + head.next=tail; + tail.prev=head; + } + + public int get(int key) { + //从缓存获取 + Node node=cache.get(key); + if(null==node){ + return -1; + } + //数据移到头结点 + moveToHead(node); + return node.value; + } + + public void put(int key, int value) { + Node node=cache.get(key); + if(null==node){ + node=new Node(key,value); + //写入新节点至头节点 + addNode(node); + cache.put(key,node); + //如果容量已满，删除尾节点 + if(size>capacity){ + //删除尾节点 + Node delNode=popTail(); + cache.remove(delNode.key); + } + }else{ + //数据更新并移到头结点 + node.value=value; + moveToHead(node); + } + } + + @Override + public String toString() { + StringBuilder sb = new StringBuilder() ; + for (Node node = head;node!=null;node=node.next){ + sb.append(node.getKey()).append(":") + .append(node.getValue()) + .append("-->"); + } + return sb.toString(); + } + public static void main(String[] args) { + LRUCache lruMap=new LRUCache(2); + lruMap.put(1,1); + lruMap.put(2,2); + lruMap.get(1); + lruMap.put(3,3); + lruMap.get(2); + lruMap.put(4,4); + lruMap.get(1); + lruMap.get(3); + lruMap.get(4); + System.out.println(lruMap.toString()); + } + +} diff --git a/src/main/java/com/algorithm/study/demo/LRUCache/LRULinked.java b/src/main/java/com/algorithm/study/demo/LRUCache/LRULinked.java index 927e8b5..46befc4 100644 --- a/src/main/java/com/algorithm/study/demo/LRUCache/LRULinked.java +++ b/src/main/java/com/algorithm/study/demo/LRUCache/LRULinked.java @@ -9,29 +9,29 @@ * @CreateDate: 2019/1/23 下午4:03 * @Version: 1.0 */ -public class LRULinked{ +public class LRULinked{ //缓存 - private final Map cacheMap = new HashMap<>(); + private final Map cacheMap = new HashMap<>(); //根节点 - private Node root; + private Node root; private int cacheSize; private int size; public LRULinked(int cacheSize){ this.cacheSize=cacheSize; } - /** * 插入头结点 * @param value */ - public void put(K key,V value){ - Node node=new Node(key,value); + public void put(int key,int value){ + cacheMap.put(key,value); + Node node=new Node(key,value); if (size==cacheSize){//容量满了删除尾节点 - Node temp=root.next; + Node temp=root.next; if (temp==null){ root=null; }else{ - Node current=root; + Node current=root; while (temp.next!=null){ current=temp; temp=temp.next; @@ -44,10 +44,13 @@ public void put(K key,V value){ root=node; size++; } - public V get(K key){ - for (Node node = root; node!=null&&!root.key.equals(key); node=node.next){ + public int get(int key){ + if (!cacheMap.containsKey(key)){ + return -1; + } + for (Node node = root; node!=null&&!root.key.equals(key); node=node.next){ if (node.next.key.equals(key)){ - Node nodeNew=new Node(node.next.key,node.next.value); + Node nodeNew=new Node(node.next.key,node.next.value); node.next=node.next.next; size--; this.put(nodeNew.key,nodeNew.value);//查找的节点放到头结点 @@ -59,36 +62,29 @@ public V get(K key){ @Override public String toString(){ StringBuilder sb=new StringBuilder(); - int j=0; - Node temp=root; - while (j{ - private K key; - private V value; - private Node next; - public Node(K key, V value){ + class Node{ + private Integer key; + private Integer value; + private Node next; + public Node(Integer key, Integer value){ this.key=key; this.value=value; } } public static void main(String[] args) { - LRULinked linked=new LRULinked<>(3); - linked.put("a","a"); - linked.put("b","b"); - linked.put("c","c"); - linked.get("b"); - linked.put("d","d"); + LRULinked linked=new LRULinked(3); + linked.put(1,2); + linked.put(2,2); + linked.put(3,3); + System.out.println(linked.get(1)); + linked.put(4,4); System.out.println(linked.size); System.out.println(linked.toString()); - System.out.println(linked.toString()); - System.out.println(linked.size); } } diff --git a/src/main/java/com/algorithm/study/demo/LRUCache/LRULinkedMap.java b/src/main/java/com/algorithm/study/demo/LRUCache/LRULinkedMap.java index 18dc367..7f84245 100644 --- a/src/main/java/com/algorithm/study/demo/LRUCache/LRULinkedMap.java +++ b/src/main/java/com/algorithm/study/demo/LRUCache/LRULinkedMap.java @@ -1,7 +1,5 @@ package com.algorithm.study.demo.LRUCache; -import java.util.ArrayList; -import java.util.Collection; import java.util.LinkedHashMap; import java.util.Map; @@ -11,33 +9,25 @@ * @CreateDate: 2018/7/12 下午8:42 * @Version: 1.0 */ -public class LRULinkedMap { +public class LRULinkedMap extends LinkedHashMap { /** * 最大缓存大小 */ - private int cacheSize; - private LinkedHashMap cacheMap ; - public LRULinkedMap(int cacheSize) { - this.cacheSize = cacheSize; - cacheMap = new LinkedHashMap(16,0.75F,true){ - @Override - protected boolean removeEldestEntry(Map.Entry eldest) { - if (cacheSize + 1 == cacheMap.size()){ - return true ; - }else { - return false ; - } - } - }; + private int CACHESIZE; + public LRULinkedMap(int cacheSize){ + // true 表示让 linkedHashMap 按照访问顺序来进行排序，最近访问的放在头部，最老访问的放在尾部。 + super(cacheSize,0.75f,true); + CACHESIZE=cacheSize; } - public void put(K key,V value){ - cacheMap.put(key,value) ; - } - public V get(K key){ - return cacheMap.get(key) ; - } - public Collection> getAll() { - return new ArrayList>(cacheMap.entrySet()); + + /** + * 删除元素条件 + * @param eldest + * @return + */ + @Override + protected boolean removeEldestEntry(Map.Entry eldest){ + return size()>CACHESIZE; } public static void main(String[] args) { LRULinkedMap map = new LRULinkedMap(4) ; @@ -47,7 +37,7 @@ public static void main(String[] args) { map.put("4",4); System.out.println(map.get("1")); map.put("5",5); - for (Map.Entry e : map.getAll()){ + for (Map.Entry e : map.entrySet()){ System.out.print(e.getKey() + " : " + e.getValue() + "\t"); } } diff --git a/src/main/java/com/algorithm/study/demo/LRUCache/LRUMap.java b/src/main/java/com/algorithm/study/demo/LRUCache/LRUMap.java deleted file mode 100644 index f0363a3..0000000 --- a/src/main/java/com/algorithm/study/demo/LRUCache/LRUMap.java +++ /dev/null @@ -1,168 +0,0 @@ -package com.algorithm.study.demo.LRUCache; - -import java.util.HashMap; -import java.util.Map; - -/** - * LRU缓存链表实现思路 - * 每次写入数据时将数据放入链表头结点。 - * 使用数据时候将数据移动到头结点。 - * 缓存数量超过阈值时移除链表尾部数据。 - * @Author: liuxun - * @CreateDate: 2018/7/12 下午6:05 - * @Version: 1.0 - */ -public class LRUMap { - private final Map cacheMap = new HashMap<>(); - /** - * 最大缓存大小 - */ - private int cacheSize; - /** - * 节点大小 - */ - private int nodeCount; - /** - * 头结点 - */ - private Node header; - /** - * 尾结点 - */ - private Node tailer; - public LRUMap(int cacheSize) { - this.cacheSize = cacheSize; - this.header=null; - this.tailer=null; - } - public void put(K key, V value) { - cacheMap.put(key, value); - //双向链表中添加结点 - addNode(key, value); - } - public V get(K key){ - Node node = getNode(key); - //移动到头结点 - moveToHead(node) ; - return cacheMap.get(key); - } - private void moveToHead(Node node){ - //如果是最后的一个节点 - if (node.next == null){ - node.tail.next=null; - tailer=node.tail; - nodeCount -- ; - } - //如果是本来就是头节点 不作处理 - if (node.tail == null){ - return ; - } - //如果处于中间节点 - if (node.tail != null && node.next != null){ - //它的上一节点指向它的下一节点 也就删除当前节点 - node.tail.next=node.next; - nodeCount -- ; - } - //最后在头部增加当前节点 - //注意这里需要重新 new 一个对象，不然原本的node 还有着下面的引用，会造成内存溢出。 - node = new Node<>(node.getKey(),node.getValue()) ; - addHead(node) ; - } - /** - * 链表查询 效率较低 - * @param key - * @return - */ - private Node getNode(K key){ - for (Node node = header;node!=null;node=node.next){ - if (node.getKey().equals(key)){ - return node ; - } - } - return null ; - } - /** - * 写入头结点 - * @param key - * @param value - */ - private void addNode(K key, V value) { - Node node = new Node<>(key, value); - //容量满了删除最后一个 - if (cacheSize == nodeCount) { - //删除尾结点 - delTail(); - } - //写入头结点 - addHead(node); - } - /** - * 添加头结点 - * - * @param node - */ - private void addHead(Node node) { - if (header==null){ - tailer=node; - }else{ - header.tail=node; - node.next=header; - } - header=node; - nodeCount++; - } - private void delTail() { - //把尾结点从缓存中删除 - cacheMap.remove(tailer.getKey()); - tailer.tail.next=null; - tailer=tailer.tail; - nodeCount--; - } - private class Node { - private K key; - private V value; - Node tail; - Node next; - public Node(K key, V value) { - this.key = key; - this.value = value; - } - public Node() { - } - public K getKey() { - return key; - } - public void setKey(K key) { - this.key = key; - } - public V getValue() { - return value; - } - public void setValue(V value) { - this.value = value; - } - } - @Override - public String toString() { - StringBuilder sb = new StringBuilder() ; - for (Node node = header;node!=null;node=node.next){ - if (node.getKey()!=null){ - sb.append(node.getKey()).append(":") - .append(node.getValue()) - .append("-->"); - } - } - return sb.toString(); - } - public static void main(String[] args) { - LRUMap lruMap=new LRUMap<>(3); - lruMap.put("1","1"); - lruMap.put("2","2"); - lruMap.put("3","3"); - lruMap.get("1"); - lruMap.put("4","4"); - System.out.println(lruMap.toString()); - System.out.println(lruMap.cacheSize); - } - -} diff --git a/src/main/java/com/algorithm/study/demo/MainTest.java b/src/main/java/com/algorithm/study/demo/MainTest.java deleted file mode 100644 index ef0534b..0000000 --- a/src/main/java/com/algorithm/study/demo/MainTest.java +++ /dev/null @@ -1,30 +0,0 @@ -package com.algorithm.study.demo; - -import java.lang.ref.PhantomReference; -import java.lang.ref.Reference; -import java.lang.ref.ReferenceQueue; - -/** - * @Author: liuxun - * @CreateDate: 2019/1/2 上午11:29 - * @Version: 1.0 - */ -public class MainTest { - public static void main(String[] args) { - Object counter = new Object(); - ReferenceQueue refQueue = new ReferenceQueue<>(); - PhantomReference p = new PhantomReference<>(counter, refQueue); - counter = null; - System.gc(); - try { - // Remove 是一个阻塞方法，可以指定 timeout，或者选择一直阻塞 - Reference ref = refQueue.remove(1000L); - if (ref != null) { - System.out.println("counter gc"); - } - } catch (InterruptedException e) { - // Handle it - } - - } -} diff --git a/src/main/java/com/algorithm/study/demo/MainTest1.java b/src/main/java/com/algorithm/study/demo/MainTest1.java new file mode 100644 index 0000000..0c3d776 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/MainTest1.java @@ -0,0 +1,36 @@ +package com.algorithm.study.demo; + +import com.alibaba.fastjson.JSON; + +import java.util.ArrayList; +import java.util.LinkedList; +import java.util.List; + +/** + * @author xun2.liu + * @title: MainTest1 + * @projectName algorithm-study + * @description: TODO + * @date 2020/6/24 17:30 + */ +public class MainTest1 { + public static void main(String[] args) { + List> lists=new ArrayList<>(); + if((lists.size() & 1)==1){ + LinkedList ls=new LinkedList<>(); + //奇数层放到队列尾部 + ls.addLast(3); + lists.add(ls); + }else{ + LinkedList ls=new LinkedList<>(); + //偶数层放到队列头部 + ls.addFirst(9); + ls.addFirst(2); + lists.add(ls); + } + + System.out.println(JSON.toJSONString(lists)); + System.out.println((2 & 1)); + + } +} diff --git a/src/main/java/com/algorithm/study/demo/MainTest2.java b/src/main/java/com/algorithm/study/demo/MainTest2.java new file mode 100644 index 0000000..0e101f0 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/MainTest2.java @@ -0,0 +1,156 @@ +package com.algorithm.study.demo; + +import java.io.*; +import java.util.*; +import java.util.concurrent.*; + +/** + * 在输入文件（/tmp/input.txt）中存储了数百个任意的文件夹绝对路径（可能重复，也可能相互包含），每个路径一行。 + * 请读入input文件内容，计算各文件夹所占用的磁盘空间大小并按占用空间由大到小的顺序进行排序。 + * 将排序后的路径及其所占用的空间大小写入到输出文件（/tmp/output.txt）。 + * 要求： + * 1、仅可使用标准JDK工具类 + * 2、请通过多线程对各统计任务进行加速 + * 3、请尽量减少全部统计任务的整体完成时间Å + * 4、请尽量保证代码规范和代码规约 + * 5、限时60分钟内完成，可任意查阅资料、博客及搜索引擎 + * 输入文件示例： + * ================== + * /home/guest + * /usr + * /usr/bin + * /var + * /root + * …… + * ================== + * 输出文件示例： + * ================== + * /root 1234567 + * /var 123456 + * /home/guest 12345 + * /usr 1234 + * /usr/bin 123 + * …… + * ================== + * @Author: liuxun + * @CreateDate: 2019/3/14 + * @Version: 1.0 + */ +public class MainTest2 { + private final static String INPUT="/tmp/input.txt"; + private final static String OUTPUT="/tmp/output.txt"; + + public static void main(String[] args) { + doReadTxt(new File(INPUT)); + System.out.println("执行结束"); + } + + public static String doReadTxt(File file) { + if (null==file || !file.exists()){ + return null; + } + BufferedReader br=null; + try { + /**创建一个线程池**/ + ExecutorService executors = new ThreadPoolExecutor(8, 8, + 60L, TimeUnit.SECONDS,new ArrayBlockingQueue(1024)); + /**读取一个文件**/ + br = new BufferedReader(new FileReader(file)); + String s = null; + Map> folders=new ConcurrentHashMap<>(); + while ((s = br.readLine()) != null) { + Future folderSizeFuture = executors.submit(new DoFolderSize(new File(s))); + folders.put(s,folderSizeFuture); + } + /**获取统计结果**/ + Map resultMap=new HashMap<>(); + for (Map.Entry> me:folders.entrySet()){ + String fileStr=me.getKey(); + Long size=me.getValue().get(); + resultMap.put(fileStr,size); + } + /**排序**/ + Map sortMap = sortMapByValue(resultMap); + for (Map.Entry fileInfo:sortMap.entrySet()){ + doOutTxt(OUTPUT,fileInfo.getKey()+" "+fileInfo.getValue()); + } + executors.shutdown();//关闭线程池 + } catch (Exception e) { + e.printStackTrace(); + }finally { + if (br!=null){ + try { + br.close(); + } catch (IOException e) { + e.printStackTrace(); + } + } + } + return null; + } + + + /** + * 异步获取文件夹大小 + */ + static class DoFolderSize implements Callable{ + private File file; + public DoFolderSize(File file){ + this.file=file; + } + @Override + public Long call() throws Exception { + return getFolderTotalSize(file); + } + } + /** + * 计算文件夹大小 + * @param file + * @return + */ + private static long getFolderTotalSize(File file) { + if (null == file) { + return 0L; + } + if (file.isFile()) { + return file.length(); + } + long total = 0; + File[] files = file.listFiles(); + for (final File f : files) { + total += getFolderTotalSize(f); + } + return total; + } + + /** + * 对map value进行排序 + * @param map + * @param + * @param + * @return + */ + public static > Map sortMapByValue(Map map) { + Map result = new LinkedHashMap<>(); + map.entrySet().stream() + .sorted((o1,o2)->o2.getValue().compareTo(o1.getValue())) + .forEach(e -> result.put(e.getKey(), e.getValue())); + return result; + } + + /** + * txt写入 + * @param path + * @param content + */ + public static void doOutTxt(String path,String content){ + try { + //构造函数中的第二个参数true表示以追加形式写文件 + FileWriter fw = new FileWriter(path,true); + fw.write(content+System.lineSeparator()); + fw.close(); + } catch (IOException e) { + e.printStackTrace(); + } + } +} diff --git a/src/main/java/com/algorithm/study/demo/OptionalTest.java b/src/main/java/com/algorithm/study/demo/OptionalTest.java new file mode 100644 index 0000000..7081aba --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/OptionalTest.java @@ -0,0 +1,21 @@ +package com.algorithm.study.demo; + +import com.algorithm.study.demo.string.TestString; + +import java.math.BigDecimal; +import java.text.NumberFormat; +import java.util.Optional; + +/** + * @title: OptionalTest + * @projectName algorithm-study + * @description: TODO + * @date 2019/8/21 10:10 + */ +public class OptionalTest { + public static void main(String[] args) { + for (int i = 5; i > 0; i--) { + System.out.println(i); + } + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/ArrayProject.java b/src/main/java/com/algorithm/study/demo/algorithm/ArraySolution.java similarity index 85% rename from src/main/java/com/algorithm/study/demo/algorithm/ArrayProject.java rename to src/main/java/com/algorithm/study/demo/algorithm/ArraySolution.java index ffb4dc9..b7455e8 100644 --- a/src/main/java/com/algorithm/study/demo/algorithm/ArrayProject.java +++ b/src/main/java/com/algorithm/study/demo/algorithm/ArraySolution.java @@ -1,12 +1,20 @@ package com.algorithm.study.demo.algorithm; -import com.sun.javadoc.SeeTag; +import java.util.HashMap; +import java.util.HashSet; +import java.util.Map; +import java.util.Set; -import java.util.*; +/** + * 数组练习题 + * @Author: liuxun + * @CreateDate: 2019/3/22 上午9:47 + * @Version: 1.0 + */ +public class ArraySolution { -public class ArrayProject { public static void main(String[] args) { - ArrayProject p=new ArrayProject(); + ArraySolution p=new ArraySolution(); System.out.println(p.containsDuplicate(new int[]{1,2,3,1})); } diff --git a/src/main/java/com/algorithm/study/demo/algorithm/DictProject.java b/src/main/java/com/algorithm/study/demo/algorithm/DictProject.java deleted file mode 100644 index 914ebcf..0000000 --- a/src/main/java/com/algorithm/study/demo/algorithm/DictProject.java +++ /dev/null @@ -1,59 +0,0 @@ -package com.algorithm.study.demo.algorithm; - -import java.util.Arrays; - -/** - * 给定一个正整数，实现一个方法来求出离该整数最近的大于自身的“换位数”。 - */ -public class DictProject { - public static void main(String[] args) { - int[] result=findNearestNumber(new int[]{1,2,3,5,4}); - for (int i=0;i0;i--){ - if (numbers[i]>numbers[i-1]){ - return i; - } - } - return 0; - } - private static void exchangeHead(int[] numbers,int index){ - int head=numbers[index-1]; - for (int i=numbers.length-1;i>0;i--){ - if (head stack=new Stack(); - ListNode tempNode=root; - while (tempNode!=null){ - stack.push(tempNode); - tempNode=tempNode.next; - } - ListNode temp=null; - while (!stack.isEmpty()){ - temp=stack.pop(); - System.out.println(temp.val); - } - - } - /**** - *单链表反转 - */ - public ListNode reversalNode(){ - //当前节点的上一个节点 - ListNode previousNode = null; - //当前节点 - ListNode currentNode = root; - //返回反转后的节点 - ListNode headNode = null; - while (currentNode != null) { - //当前节点的下一个节点 - ListNode nextNode = currentNode.next; - if (nextNode == null) { - headNode = currentNode; - } - currentNode.next = previousNode; - previousNode = currentNode; - currentNode = nextNode; - System.out.println(); - } - return headNode; - } - /** - * 链表中环检测 - * *合并两个有序链表 - * * 删除链表倒数第N个节点 - * * 求链表的中间节点 - * @param args - */ - - - public static void main(String[] args) { - Jianzhi02 listnode=new Jianzhi02(); - listnode.add(4); - listnode.add(3); - listnode.add(2); - listnode.add(1); - System.out.println("size:"+listnode.size()); - listnode.printListInverselyUsingIteration(); - listnode.reversalNode(); - } -} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/Solution.java b/src/main/java/com/algorithm/study/demo/algorithm/Solution.java new file mode 100644 index 0000000..4d67495 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/Solution.java @@ -0,0 +1,86 @@ +package com.algorithm.study.demo.algorithm; + +import com.alibaba.fastjson.JSON; + +import java.util.*; + +/** + * 算法面试题 + * @Author: liuxun + * @CreateDate: 2019/2/13 下午3:11 + * @Version: 1.0 + */ +public class Solution { + + /** + * 整数反转 + * 给出一个 32 位的有符号整数，你需要将这个整数中每位上的数字进行反转。 + * @param x + * @return + */ + public static int reverse(int x) { + long result = 0; + while (x != 0) { + result = result * 10 + x % 10; + x /= 10; + } + if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) + result = 0; + return (int)result; + } + + /** + * 求一个字符串的最大不重复子串 + */ + public static int lengthOfLongestSubstring(String s) { + int n = s.length(); + Set set = new HashSet<>(); + int ans = 0, i = 0, j = 0; + while (i < n && j < n) { + // try to extend the range [i, j] + if (!set.contains(s.charAt(j))){ + set.add(s.charAt(j++)); + ans = Math.max(ans, j - i); + } + else { + //删除重复的字符 + set.remove(s.charAt(i++)); + } + } + return ans; + } + + /** + * 求2个有序数组的有序交集 + * @param args + */ + public static void arrayIntersection(int[] a,int[] b){ + int lena=a.length; + int lenb=b.length; + if (lena==0 || lenb==0){ + System.out.println("无数据交集"); + return; + } + int i=0,j=0; + List result=new ArrayList<>(); + while (ib[j]){ + j++; + }else{ + result.add(a[i]); + i++; + j++; + } + } + System.out.println("交集为："+ JSON.toJSONString(result)); + } + public static void main(String[] args) { + System.out.println(lengthOfLongestSubstring("abaa")); + int[] a = { 2, 3, 4, 4, 4, 4, 7, 8, 8, 8, 8, 9, 100, 130, 150, 160 }; + int[] b = { 4, 6, 7, 7, 7, 7, 8, 8, 9, 10, 100, 130, 130, 140, 150 }; + arrayIntersection(a,b); + } + +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/SortProject.java b/src/main/java/com/algorithm/study/demo/algorithm/SortProject.java index 0f0c453..ec90f5d 100644 --- a/src/main/java/com/algorithm/study/demo/algorithm/SortProject.java +++ b/src/main/java/com/algorithm/study/demo/algorithm/SortProject.java @@ -61,17 +61,17 @@ public static void performanceTest(int len) throws Exception{ * 时间复杂度为O(n²) * 空间复杂度为O(1) */ - private static void maopaoSort(int score[]){ - System.out.println("排序前："+ JSON.toJSONString(score)); - boolean flag=true;//数据发生了交换才继续冒泡 - for (int i = 1; i < score.length && flag; i++){ //最多做n-1趟排序 - flag=false; - for(int j = 0 ;j < score.length - i; j++){ //对当前无序区间score[0......length-i-1]进行排序(j的范围很关键，这个范围是在逐步缩小的) - if(score[j] > score[j + 1]){ //把大或者小的值交换到后面 - int temp = score[j]; - score[j] = score[j + 1]; - score[j + 1] = temp; - flag=true;//发生了数据交换 + private static void maopaoSort(int score[]){ + System.out.println("排序前："+ JSON.toJSONString(score)); + boolean flag=true;//数据发生了交换才继续冒泡 + for (int i = 1; i < score.length && flag; i++){ //最多做n-1趟排序 + flag=false; + for(int j = 0 ;j < score.length - i; j++){ //对当前无序区间score[0......length-i-1]进行排序(j的范围很关键，这个范围是在逐步缩小的) + if(score[j] > score[j + 1]){ //把大或者小的值交换到后面 + int temp = score[j]; + score[j] = score[j + 1]; + score[j + 1] = temp; + flag=true;//发生了数据交换 } } } @@ -172,6 +172,7 @@ public static void quickSort(int[] a,int low,int high) { int hi = high; if (lo >= hi) return; + //11,1,15,30,40 //确定指针方向的逻辑变量 boolean transfer=true; while (lo != hi) { @@ -250,26 +251,31 @@ public static void quick(int[] a2) { * @param b 有序数组2 * @return 合并之后的有序数组； */ - public static int[] merge(int[] a, int[] b){ - int result[] = new int[a.length+b.length]; - int i=0,j=0,k=0; - while(i @@ -283,18 +289,19 @@ public static int[] merge(int[] a, int[] b){ * 稳定排序算法 * 不是原地排序算法 */ - public static int[] mergeSort(int[] a){ - if(a.length==1){ - return a; + public static int[] mSort(int[] a,int low,int high){ + int mid = (low+high)/2; + if(low objects = new HashSet<>(); + for (String w:words){ + StringBuilder sb=new StringBuilder(); + char[] chars = w.toCharArray(); + for (char aChar : chars) { + sb.append(m[aChar-'a']); + } + System.out.println("密码为："+sb); + objects.add(sb.toString()); + } + return objects.size(); + } + + public static void main(String[] args) { +// System.out.println(StringSolution.indexOf("ab","asdkfjasldjfab")); +// System.out.println("asdkfjasldjfab".indexOf("ab")); + +// System.out.println(StringSolution.toLowerCase("ccccccccccBB")); + System.out.println(StringSolution.reverseString("abc".toCharArray())); + System.out.println(StringSolution.uniqueMorseRepresentations(new String[]{"abc","cba"})); + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/Suanfa50.java b/src/main/java/com/algorithm/study/demo/algorithm/Suanfa50.java deleted file mode 100644 index c06afba..0000000 --- a/src/main/java/com/algorithm/study/demo/algorithm/Suanfa50.java +++ /dev/null @@ -1,33 +0,0 @@ -package com.algorithm.study.demo.algorithm; - -import java.util.Arrays; -import java.util.BitSet; - -/** - * @Author: liuxun - * @CreateDate: 2018/12/7 下午1:35 - * @Version: 1.0 - */ -public class Suanfa50 { - public static void main(String[] args) { - printMissingNumber(new int[]{1, 3, 6}, 6); - } - private static void printMissingNumber(int[] numbers, int count) { - int missingCount = count - numbers.length; - BitSet bitSet = new BitSet(count); - - for (int number : numbers) { - bitSet.set(number - 1); - } - - System.out.printf("Missing numbers in integer array %s, with total number %d is %n", - Arrays.toString(numbers), count); - int lastMissingIndex = 0; - - for (int i = 0; i < missingCount; i++) { - lastMissingIndex = bitSet.nextClearBit(lastMissingIndex); - System.out.println(++lastMissingIndex); - } - - } -} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/TanxinSolution.java b/src/main/java/com/algorithm/study/demo/algorithm/TanxinSolution.java new file mode 100644 index 0000000..c126966 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/TanxinSolution.java @@ -0,0 +1,30 @@ +package com.algorithm.study.demo.algorithm; + +/** + * 贪心算法 + * @Author: liuxun + * @CreateDate: 2019/3/22 上午11:01 + * @Version: 1.0 + */ +public class TanxinSolution { + public static void main(String[] args) { + TanxinSolution.greedyGiveMoney(10); + } + /** + * 钱币找零问题 + *假设1元、2元、5元、10元、20元、50元、100元的纸币，张数不限制，现在要用来支付K元，至少要多少张纸币？ + * @param money the money + */ + public static void greedyGiveMoney(int money) { + System.out.println("需要找零: " + money); + int[] moneyLevel = {1, 5, 10, 20, 50, 100}; + for (int i=moneyLevel.length-1;i>=0;i--){ + int num=money/moneyLevel[i];//张数 + int mod=money%moneyLevel[i];//剩余的钱 + money=mod; + if (num>0){ + System.out.println("最少需要"+num+"张"+moneyLevel[i]+"元的纸币"); + } + } + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/ZumaProject.java b/src/main/java/com/algorithm/study/demo/algorithm/ZumaProject.java deleted file mode 100644 index dfdd19f..0000000 --- a/src/main/java/com/algorithm/study/demo/algorithm/ZumaProject.java +++ /dev/null @@ -1,96 +0,0 @@ -package com.algorithm.study.demo.algorithm; - -import com.sun.deploy.util.StringUtils; - -import java.util.*; -import java.util.regex.Matcher; -import java.util.regex.Pattern; - -/** - * 祖玛消除算法 - * @Author: liuxun - * @CreateDate: 2018/11/6 下午3:58 - * @Version: 1.0 - */ -public class ZumaProject { - public static void main(String[] args) { - String[] is = new String[50]; - Random ra =new Random(); - for(int i=0;i<50;i++){ - is[i]=ra.nextInt(3)+""; - } - //把输入的数据转换成字符串。这样可以利用正则表达式去查找重复数字 - String input=StringUtils.join(Arrays.asList(is),""); - System.out.println("初始化数据："+input); - String newInput=matcher(input); - System.out.println("去掉初始化数据连续出现三次级以上数字后的结果："+newInput); - - Scanner sc = new Scanner(System.in); - System.out.println("输入需要消除的数字下标:"); - int index=sc.nextInt(); - String findInput=find(index,newInput); - System.out.println("消除后的结果："+findInput); - - - } - public static String matcher(String input) - { - //创建一个List 用于存放重复3次或者以上的的数字 - List list = new ArrayList(); - //创建匹配的模式 - Pattern pattern = Pattern.compile("(\\w)\\1{2,}"); - //匹配器 - Matcher matcher = pattern.matcher(input); - //查找与该模式匹配的子序列。从input 里面 查找出 与 此模式 "(\w)\1{2,}" 相匹配的 子序列。如果存在，返回true，如果不存在，返回false. - while (matcher.find()) - { - //返回匹配的子序列，并加入到list里面。 - list.add(matcher.group()); - } - if (list.isEmpty()){ - System.out.println("当前数据没有出现连续三次或以上的数字"); - return input; - } - System.out.println("当前连续三次或以上出现的数字为："+list); - //找到连续重复的字符,加入到数组中。 - String[] strings = list.toArray(new String[0]); - //找出连续并且重复的子序列。并且把这些连续重复的子序列用空字符串替换。 - for(int i=0 ;i<=strings.length-1;i++){ - if(strings[i].length()>1){ -// System.out.println(strings[i]); - input=input.replace(strings[i],""); -// System.out.println(input); - } - } - //返回把连续重复的字符赐除掉的字符序列。 - return input; - } - - /** - * 删除字符串指定index的元素 - * @param index - * @param Str - * @return - */ - public static String removeChar(int index,String Str) { - Str = Str.substring(0, index) + Str.substring(index + 1, Str.length());//substring的取值范围是:[,) - return Str; - } - /** - * 用户需要消除的数字下标以及数字 - * 思路就是删掉用户查找的那个数字然后返回新的字符串 - * 比如{11212}，用户传入下标为2的数字2返回"1112" - * 然后把字符串继续重复调用matcher即可 - * @param index - * @return - */ - public static String find(int index,String array){ - String newArr=removeChar(index,array); - String matcherArr = matcher(newArr); - if (newArr.equals(matcherArr)){ - //如果用户输入数据后没有发现连续三次级以上的数据就返回原始数据 - return array; - } - return matcherArr; - } -} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/dynamicprogramming/demo1.java b/src/main/java/com/algorithm/study/demo/algorithm/dynamicprogramming/demo1.java new file mode 100644 index 0000000..31b98a5 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/dynamicprogramming/demo1.java @@ -0,0 +1,88 @@ +package com.algorithm.study.demo.algorithm.dynamicprogramming; + +/** + * @author xun2.liu + * @title: demo1 + * @projectName algorithm-study + * @description: 动态规划-斐波那契数列 + * @date 2020/3/18 14:04 + */ +public class demo1 { + public static void main(String[] args) { + System.out.println(fib(10)); + System.out.println(fib2(10)); + System.out.println(fib3(10)); + System.out.println(fib4(10)); + } + + /** + * 重叠子问题，子问题个数为 O(2^n)。性能非常低。 + * @param n + * @return + */ + private static int fib(int n){ + if(n==1 || n==2){ + return 1; + } + return fib(n-1)+fib(n-2); + } + + /** + * 带备忘录的递归解法,解决重叠子问题。 + * 子问题个数为 O(n),时间复杂度是 O(n) + * @param n + * @return + */ + private static int fib2(int n){ + if(n==1 || n==2){ + return 1; + } + int[] memo=new int[n+1]; + return helper(memo,n); + } + private static int helper(int[] memo,int n){ + if(n==1 || n==2){ + return 1; + } + if (memo[n]!=0) { + return memo[n]; + } + return memo[n]=helper(memo,n-1)+helper(memo,n-2); + } + + /** + * DP table 数组的迭代解法 + * 空间复杂度降为 O(N) + *自底向上计算斐波那契数列 + * @param n + * @return + */ + private static int fib3(int n){ + int[] temp=new int[n+1]; + temp[1]=temp[2]=1; + for (int i=3;i<=n;i++){ + temp[i]=temp[i-1]+temp[i-2]; + } + return temp[n]; + } + + /** + *DP table 数组的迭代解法优化 + *空间复杂度降为 O(1) + *自底向上计算斐波那契数列 + * @param n + * @return + */ + private static int fib4(int n){ + if (n==1 || n==2){ + return 1; + } + int prev=1;int curr=1; + for (int i=3;i<=n;i++){ + int temp=prev+curr; + prev=curr; + curr=temp; + } + return curr; + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/greedyalgorithm/MoneyBusi.java b/src/main/java/com/algorithm/study/demo/algorithm/greedyalgorithm/MoneyBusi.java new file mode 100644 index 0000000..e402baa --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/greedyalgorithm/MoneyBusi.java @@ -0,0 +1,25 @@ +package com.algorithm.study.demo.algorithm.greedyalgorithm; + +import lombok.*; + +/** + * @title: MoneyBusi + * @projectName algorithm-study + * @description: 零钱支付 + * @date 2019/11/14 17:19 + */ +@Getter +@Setter +@ToString +@AllArgsConstructor +@NoArgsConstructor +public class MoneyBusi { + /** 面值 */ + private String value; + + /** 张数 */ + private int num; + + /** 金额 */ + private int memory; +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/greedyalgorithm/Solutions.java b/src/main/java/com/algorithm/study/demo/algorithm/greedyalgorithm/Solutions.java new file mode 100644 index 0000000..761a235 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/greedyalgorithm/Solutions.java @@ -0,0 +1,100 @@ +package com.algorithm.study.demo.algorithm.greedyalgorithm; + +import java.util.ArrayList; +import java.util.List; +import java.util.PriorityQueue; + +/** + * @author liuxun + * @title: Solutions + * @projectName algorithm-study + * @description: 零钱支付问题 + * 假设我们有 1 元、2 元、5 元、10 元、20 元、50 元、100 元这些面额的纸币， + * 它们的张数分别是 c1、c2、c5、c10、c20、c50、c100。 + * 我们现在要用这些钱来支付 K 元，最少要用多少张纸币呢？ + * 先用面值最大的来支付，如果不够，就继续用更小一点面值的，以此类推，最后剩下的用 1 元来补齐。 + * @date 2019/11/14 17:20 + */ +public class Solutions { + /** + * 用于存储金额的信息,根据金额从大到小排序 + */ + public PriorityQueue moneyQueue = + new PriorityQueue<>( + (o1, o2) -> { + if (o1.getMemory() < o2.getMemory()) { + return 1; + } else if (o1.getMemory() > o2.getMemory()) { + return -1; + } + return 0; + }); + + /** + * 添加金额信息 + * @param value 面值信息 + * @param num 张数 + * @param memory 金额值 + */ + public void addMemoryInfo(String value, int num, int memory) { + moneyQueue.offer(new MoneyBusi(value, num, memory)); + } + + /** + * 计算找零钱的问题 + * + * @param money 找零的金额信息 + * @return 找零钱的信息 + */ + public List looseChange(int money) { + + List resultMemory = new ArrayList<>(); + + List moreMemory = new ArrayList<>(); + + int surplus = money; + + while (surplus > 0) { + //返回队列头部元素 + MoneyBusi busi = moneyQueue.peek(); + if (null != busi) { + System.out.println("当前金额："+busi.getMemory()); + if (busi.getMemory() <= surplus) { + busi = moneyQueue.poll(); + surplus = surplus - busi.getMemory(); + + MoneyBusi busiNew = new MoneyBusi(busi.getValue(), 1, busi.getMemory()); + resultMemory.add(busiNew); + + busi.setNum(busi.getNum() - 1); + + if (busi.getNum() > 0) { + moneyQueue.offer(busi); + } + } else { + moreMemory.add(moneyQueue.poll()); + } + } else { + break; + } + } + moneyQueue.addAll(moreMemory); + return resultMemory; + } + + public static void main(String[] args) { + Solutions instance = new Solutions(); + instance.addMemoryInfo("100元", 2, 100); + instance.addMemoryInfo("50元", 2, 50); + instance.addMemoryInfo("20元", 2, 20); + instance.addMemoryInfo("10元", 2, 10); + instance.addMemoryInfo("5元", 2, 5); + instance.addMemoryInfo("2元", 2, 2); + instance.addMemoryInfo("1元", 5, 1); + System.out.println(instance.moneyQueue); + List list = instance.looseChange(332); + for (MoneyBusi busi : list) { + System.out.println(busi); + } + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/huffman/DataOutputStreamHuffman.java b/src/main/java/com/algorithm/study/demo/algorithm/huffman/DataOutputStreamHuffman.java new file mode 100644 index 0000000..1d622e4 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/huffman/DataOutputStreamHuffman.java @@ -0,0 +1,59 @@ +package com.algorithm.study.demo.algorithm.huffman; +import java.io.FileNotFoundException; +import java.io.FileOutputStream; +import java.io.IOException; + +/** + * @author xun2.liu + * @title: DataOutputStreamHuffman + * @projectName algorithm-study + * @description: TODO + * @date 2019/11/19 17:03 + */ +public class DataOutputStreamHuffman { + + public static final DataOutputStreamHuffman OUTPUT = new DataOutputStreamHuffman(); + + public static final String path = "D:\\java\\test\\datastruct\\hoffman\\"; + + public void outtoFile(byte[] value) { + FileOutputStream output = null; + try { + output = new FileOutputStream(path + "src.file"); + output.write(value); + } catch (FileNotFoundException e) { + e.printStackTrace(); + } catch (IOException e) { + e.printStackTrace(); + } finally { + if (null != output) { + try { + output.close(); + } catch (IOException e) { + e.printStackTrace(); + } + } + } + } + + public void outHuffmantoFile(byte[] value) { + FileOutputStream output = null; + try { + output = new FileOutputStream(path + "out.huff"); + output.write(value); + } catch (FileNotFoundException e) { + e.printStackTrace(); + } catch (IOException e) { + e.printStackTrace(); + } finally { + if (null != output) { + try { + output.close(); + } catch (IOException e) { + e.printStackTrace(); + } + } + } + } + +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/huffman/HuffmanCode.java b/src/main/java/com/algorithm/study/demo/algorithm/huffman/HuffmanCode.java new file mode 100644 index 0000000..6de61a1 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/huffman/HuffmanCode.java @@ -0,0 +1,193 @@ +package com.algorithm.study.demo.algorithm.huffman; + +import java.util.*; +/** + * @author xun2.liu + * @title: HuffmanCode + * @projectName algorithm-study + * @description: Huffman编码 + * @date 2019/11/19 17:03 + */ +public class HuffmanCode { + /** 根节点信息，根据编码后设置根节点信息 */ + public CodeNode root; + + /** 节点信息 */ + public class CodeNode { + /** 编码存储的数据信息 */ + public char data; + + /** 字符出现的频率 */ + public int frequence; + + /** 霍夫漫编码左节点 */ + public CodeNode left; + + /** 霍夫漫编码右节点 */ + public CodeNode right; + + /** 父节点信息 */ + public CodeNode parent; + + /** 标识是否为计算添加节点 */ + public boolean addNode; + + @Override + public String toString() { + final StringBuilder sb = new StringBuilder("CodeNode{"); + sb.append("data=").append(data); + sb.append(", frequence=").append(frequence); + sb.append('}'); + return sb.toString(); + } + } + /** + * 编码，形成每个字符的霍夫漫编码 + * + * @param map 统计信息 + * @return 结果编码后的信息 + */ + public Map getHuffManCode(Map map) { + if (null != map && !map.isEmpty()) { + // 使用小顶堆来进行数据的存储 + PriorityQueue nodePriQueue = + new PriorityQueue<>( + map.size(), + (o1, o2) -> { + if (o1.frequence > o2.frequence) { + return 1; + } else if (o1.frequence < o2.frequence) { + return -1; + } + return 0; + }); + + CodeNode nodeTmp = null; + + // 1,将数据放入小顶堆中 + for (Map.Entry item : map.entrySet()) { + + nodeTmp = new CodeNode(); + nodeTmp.data = item.getKey(); + nodeTmp.frequence = item.getValue(); + + nodePriQueue.offer(nodeTmp); + } + + int queueSize = nodePriQueue.size(); + + // 将统计数据编译成霍夫漫编码树信息 + for (int i = 1; i < queueSize; i++) { + CodeNode left = nodePriQueue.poll(); + CodeNode right = nodePriQueue.poll(); + + CodeNode sumNode = new CodeNode(); + sumNode.frequence = left.frequence + right.frequence; + sumNode.data = (char) ((int) left.data + (int) right.data); + sumNode.addNode = true; + + sumNode.left = left; + sumNode.right = right; + + left.parent = sumNode; + right.parent = sumNode; + + nodePriQueue.offer(sumNode); + } + + // 构建完成 + root = nodePriQueue.poll(); + + // 构建霍夫漫编码 + Map result = this.builderCode(root); + + return result; + } + + return null; + } + + public Map builderCode(CodeNode root) { + + Map result = new HashMap<>(); + + StringBuilder code = new StringBuilder(); + + this.buildCode(code, root, result); + + return result; + } + + /** + * 进行霍夫温编码的操作，此处使用递归来实现 + * + * @param code 主串信息 + * @param node 霍夫漫编码树信息 + * @param result 存储的结果节点信息 + */ + private void buildCode(StringBuilder code, CodeNode node, Map result) { + if (null == node) { + return; + } + + if (!node.addNode) { + result.put(node.data, code.toString()); + } + + if (node.left != null) { + code.append("0"); + this.buildCode(code, node.left, result); + code.deleteCharAt(code.length() - 1); + } + + if (node.right != null) { + code.append("1"); + this.buildCode(code, node.right, result); + code.deleteCharAt(code.length() - 1); + } + } + + public String parseHuffman2(String src, Map huffman) { + StringBuilder out = new StringBuilder(); + + char[] hufSrcs = src.toCharArray(); + + for (char hufs : hufSrcs) { + out.append(huffman.get(hufs)); + } + + return out.toString(); + } + + /** + * 进行霍夫漫的解码操作 + * + * @param hufStr + * @param root + * @return + */ + public String decodeHuffman(String hufStr, CodeNode root) { + char[] hubs = hufStr.toCharArray(); + + int index = 0; + + StringBuilder resultMsg = new StringBuilder(); + + while (index < hubs.length) { + CodeNode node = root; + + do { + if (hubs[index] == '0') { + node = node.left; + } else if (hubs[index] == '1') { + node = node.right; + } + index++; + } while (index < hubs.length && (node.left != null || node.right != null)); + + resultMsg.append(node.data); + } + + return resultMsg.toString(); + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/huffman/StrProc.java b/src/main/java/com/algorithm/study/demo/algorithm/huffman/StrProc.java new file mode 100644 index 0000000..cf753bd --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/huffman/StrProc.java @@ -0,0 +1,42 @@ +package com.algorithm.study.demo.algorithm.huffman; + +import java.util.HashMap; +import java.util.Map; +/** + * @author xun2.liu + * @title: StrProc + * @projectName algorithm-study + * @description: 对字符进行统计 + * @date 2019/11/19 17:07 + */ +public class StrProc { + /** + * 对字符进行统计操作 + * + * @param str + * @return + */ + public static Map countCharset(String str) { + + if (null != str && str.length() > 0) { + + Map result = new HashMap<>(); + + char[] strChars = str.toCharArray(); + + Integer value = null; + for (int i = 0; i < strChars.length; i++) { + value = result.get(strChars[i]); + + if (value == null) { + result.put(strChars[i], 1); + } else { + result.put(strChars[i], value + 1); + } + } + + return result; + } + return null; + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/huffman/TestHuffmanCode.java b/src/main/java/com/algorithm/study/demo/algorithm/huffman/TestHuffmanCode.java new file mode 100644 index 0000000..af4f299 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/huffman/TestHuffmanCode.java @@ -0,0 +1,41 @@ +package com.algorithm.study.demo.algorithm.huffman; + + +import java.util.Map; +import java.util.concurrent.ThreadLocalRandom; +/** + * @author xun2.liu + * @title: TestHuffmanCode + * @projectName algorithm-study + * @description: huffman编码测试 + * @date 2019/11/19 17:08 + */ +public class TestHuffmanCode { + public static void main(String[] args) { +// for (int i = 0; i < 5; i++) { + int valueRand = ThreadLocalRandom.current().nextInt(1, 50); + StringBuilder msg = new StringBuilder(); + for (int j = 0; j < valueRand; j++) { + msg.append((char) ThreadLocalRandom.current().nextInt(65, 122)); + } + + String src = "aaaaaaabbcceed" + msg.toString(); + System.out.println("原始：" + src); + + Map conMap = StrProc.countCharset(src); + + System.out.println("字符频率统计："+conMap); + + HuffmanCode instance = new HuffmanCode(); + Map huffCode = instance.getHuffManCode(conMap); + System.out.println("huffcode字符编码映射："+huffCode); + + String hufOutValue = instance.parseHuffman2(src, huffCode); + System.out.println("最终编码："+hufOutValue); + + String deValue = instance.decodeHuffman(hufOutValue, instance.root); + System.out.println("解压结果：" + deValue); + + System.out.println("--------------------------------------------------------------------------------"); + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution.java b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution.java new file mode 100644 index 0000000..97af0e9 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution.java @@ -0,0 +1,78 @@ +package com.algorithm.study.demo.algorithm.leetcode; + +import com.algorithm.study.demo.LRUCache.LRUCache; + +/** + * @author xun2.liu + * @title: Solution + * @projectName algorithm-study + * @description: + * 给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的， + * 并且它们的每个节点只能存储 一位 数字。 + * + * 如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。 + * + * 您可以假设除了数字 0 之外，这两个数都不会以 0 开头。 + * 示例： + * + * 输入：(2 -> 4 -> 3) + (5 -> 6 -> 4) + * 输出：7 -> 0 -> 8 + * 原因：342 + 465 = 807 + * + * 来源：力扣（LeetCode） + * 链接：https://leetcode-cn.com/problems/add-two-numbers + * @date 2020/5/9 14:42 + */ +public class Solution { + static class ListNode{ + private int val; + private ListNode next; + private ListNode(int val){ + this.val=val; + } + + } + + public static ListNode addTwoNumbers(ListNode l1, ListNode l2) { + //返回的结果,初始化 + ListNode result=new ListNode(0); + //j结果游标 + ListNode curr=result; + //满十进1，存储进位 + int carry=0; + while(l1!=null || l2!=null){ + int p1=l1==null?0:l1.val; + int p2=l2==null?0:l2.val; + //计算当前两数相加后的值 + int sum=p1+p2+carry; + //计算相加后的值的进位 + carry=sum/10; + //存储当前相加后的值除以10的余数 + curr.next=new ListNode(sum%10); + //游标指向下个节点 + curr=curr.next; + + if (l1!=null){ + l1=l1.next; + } + if (l2!=null){ + l2=l2.next; + } + } + if (carry>0){ + curr.next=new ListNode(carry); + } + return result.next; + } + public static void main(String[] args) { + ListNode a=new ListNode(5); + ListNode b=new ListNode(5); + a.next=b; + + + ListNode result = addTwoNumbers(a, a); + for (ListNode node=result;node!=null;node=node.next){ + System.out.println(node.val); + } + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution10.java b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution10.java new file mode 100644 index 0000000..21e3809 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution10.java @@ -0,0 +1,58 @@ +package com.algorithm.study.demo.algorithm.leetcode; + +import com.alibaba.fastjson.JSON; + +import java.util.ArrayList; +import java.util.List; + +/** + * @author xun2.liu + * @title: Solution10 + * @projectName algorithm-study + * @description: 给定一个字符串，你的任务是计算这个字符串中有多少个回文子串。 + * + * 具有不同开始位置或结束位置的子串，即使是由相同的字符组成，也会被计为是不同的子串。 + * + * 示例 1: + * + * 输入: "abc" + * 输出: 3 + * 解释: 三个回文子串: "a", "b", "c". + * 示例 2: + * + * 输入: "aaa" + * 输出: 6 + * 说明: 6个回文子串: "a", "a", "a", "aa", "aa", "aaa". + * + * 来源：力扣（LeetCode） + * 链接：https://leetcode-cn.com/problems/palindromic-substrings + * @date 2020/5/21 15:34 + */ +public class Solution10 { + //通过中心扩散法查找回文字符串 + public static void huiwen(String s,int l,int r,List filter){ + while(l>=0 && r filter=new ArrayList(); + for(int i=0;ilen){ + return -1; + } + if (len==n && haystack.equals(needle)){ + return 0; + } + for (int i=0;i2->4, 1->3->4 + * 输出：1->1->2->3->4->4 + * + * 来源：力扣（LeetCode） + * 链接：https://leetcode-cn.com/problems/merge-two-sorted-lists + */ +public class Solution14 { + public static class ListNode { + int val; + ListNode next; + ListNode(int x) { val = x; } + } + public static ListNode mergeTwoLists(ListNode l1, ListNode l2) { + //需要返回排序好的节点 + ListNode result=new ListNode(-1); + //哨兵节点 + ListNode prev=result; + //首先遍历两个链表比较大小如果l1比l2小。l1往前走否则l2往前走。并且把值小的节点赋值给prev.next。 + while(l1!=null && l2!=null){ + if(l1.val map= new HashMap<>(); + for(int i=0;i queue=new LinkedList<>(); + //把当前的第一层添加至队列中 + queue.offer(root); + //默认深度为0 + int depth=0; + while (queue.size()>0){ + //获取当前层的节点数量 + int size = queue.size(); + //遍历当前层的节点 + for (int i = 0; i < size; i++) { + //弹出当前层的节点。获取节点下一层的节点 + TreeNode head = queue.poll(); + if (head.left!=null){ + queue.offer(head.left); + } + if (head.right!=null){ + queue.offer(head.right); + } + } + //当前层遍历结束后。树的深度+1 + depth++; + } + return depth; + + } + public static void main(String[] args) { + TreeNode a=new TreeNode(3); + TreeNode b=new TreeNode(9); + TreeNode c=new TreeNode(20); + TreeNode d=new TreeNode(15); + TreeNode r=new TreeNode(7); + a.left=b; + a.right=c; + c.left=d; + c.right=r; + System.out.println(maxDepth(a)); + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution19.java b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution19.java new file mode 100644 index 0000000..df36d40 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution19.java @@ -0,0 +1,79 @@ +package com.algorithm.study.demo.algorithm.leetcode; + +import java.util.ArrayList; +import java.util.List; +import java.util.Stack; + +/** + * @author xun2.liu + * @title: Solution19 + * @projectName algorithm-study + * @description: 给定一个二叉树，返回它的中序 遍历。 + * + * 示例: + * + * 输入: [1,null,2,3] + * 1 + * \ + * 2 + * / + * 3 + * + * 输出: [1,3,2] + * + * 来源：力扣（LeetCode） + * 链接：https://leetcode-cn.com/problems/binary-tree-inorder-traversal + * @date 2020/5/29 18:31 + */ +public class Solution19 { + public static class TreeNode { + int val; + TreeNode left; + TreeNode right; + TreeNode(int x) { val = x; } + } + + /** + * 前序遍历:打印-左-右 + * 中序遍历:左-打印-右 + * 后序遍历:左-右-打印 + * @param root + * @return + */ + public static List inorderTraversal(TreeNode root) { + List result=new ArrayList(); + Stack stack=new Stack(); + TreeNode curr=root; + while(curr!=null || !stack.isEmpty()){ + //首先遍历左子节点 + if (curr!=null){ + //不断往左子树方向走，每走一次就将当前节点保存到栈中 + //这是模拟递归的调用 + stack.push(curr); + curr=curr.left; + }else{ + //当前节点为空，说明左边走到头了，从栈中弹出节点并保存 + //然后转向右边节点，继续上面整个过程 + curr= stack.pop(); + result.add(curr.val); + curr=curr.right; + } + } + return result; + } + public static void main(String[] args) { + TreeNode a=new TreeNode(3); + TreeNode b=new TreeNode(9); + TreeNode c=new TreeNode(20); + TreeNode d=new TreeNode(15); + TreeNode r=new TreeNode(2); + a.left=b; + a.right=c; + c.left=d; + c.right=r; + List integers = inorderTraversal(a); + for (Integer integer : integers) { + System.out.println(integer); + } + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution2.java b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution2.java new file mode 100644 index 0000000..5eae01b --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution2.java @@ -0,0 +1,58 @@ +package com.algorithm.study.demo.algorithm.leetcode; + +import java.util.HashMap; +import java.util.Map; + +/** + * @author xun2.liu + * @title: Solution2 + * @projectName algorithm-study + * @description: + * 给定一个字符串，请你找出其中不含有重复字符的 最长子串 的长度。 + * 示例 1: + * + * 输入: "abcabcbb" + * 输出: 3 + * 解释: 因为无重复字符的最长子串是 "abc"，所以其长度为 3。 + * + * 来源：力扣（LeetCode） + * 链接：https://leetcode-cn.com/problems/longest-substring-without-repeating-characters + * @date 2020/5/13 15:51 + */ +public class Solution2 { + /** + * 使用滑动窗口 + * 定义一个 map 数据结构存储 (k, v)，其中 key 值为字符，value 值为字符位置 +1，加 1 表示从字符位置后一个才开始不重复 + * 我们定义不重复子串的开始位置为 start，结束位置为 end + * 随着 end 不断遍历向后，会遇到与 [start, end] 区间内字符相同的情况，此时将字符作为 key 值，获取其 value 值，并更新 start，此时 [start, end] + * 区间内不存在重复字符 + * 无论是否更新 start，都会更新其 map 数据结构和结果 ans。 + * 时间复杂度：O(n)O(n) + * @param s + * @return + */ + public static int lengthOfLongestSubstring(String s) { + //最长子串 的长度 + int resultLen=0; + //key为字符 value为end + Map map= new HashMap<>(); + //初始化起始位置和结束位置 + int start=0; + for(int end=0;end='0' && str.charAt(i)<='9') { + //'0'的ASCII码是48，'1'的是49，这么一减就从就可以得到真正的整数值 + int tmp = str.charAt(i)-48; + //判断是否大于 最大32位整数 + if(!is_negative &&(res>Integer.MAX_VALUE ||(res==Integer.MAX_VALUE && tmp>=7))) { + return Integer.MAX_VALUE; + } + //判断是否小于 最小32位整数 + if(is_negative &&(-res<-Integer.MAX_VALUE || (-res==-Integer.MAX_VALUE && tmp>=8))) { + return -Integer.MAX_VALUE; + } + res = res*10 + tmp; + ++i; + } + //如果有负号标记则返回负数 + if(is_negative) { + return -res; + } + return res; + } + + public static void main(String[] args) { + System.out.println(myAtoi("-42")); + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution22.java b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution22.java new file mode 100644 index 0000000..836babf --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution22.java @@ -0,0 +1,46 @@ +package com.algorithm.study.demo.algorithm.leetcode; + +/** + * @author xun2.liu + * @title: Solution22 + * @projectName algorithm-study + * @description: 字符串压缩。利用字符重复出现的次数，编写一种方法，实现基本的字符串压缩功能。 + * 比如，字符串aabcccccaaa会变为a2b1c5a3。若“压缩”后的字符串没有变短，则返回原先的字符串。 + * 你可以假设字符串中只包含大小写英文字母（a至z）。 + * 示例1: + * + * 输入："aabcccccaaa" + * 输出："a2b1c5a3" + * + * 来源：力扣（LeetCode） + * 链接：https://leetcode-cn.com/problems/compress-string-lcci + * @date 2020/6/3 20:07 + */ +public class Solution22 { + public static String compressString(String S) { + //慢指针 + int i=0; + int len=S.length(); + //压缩后的字符串 + StringBuilder sb=new StringBuilder(); + while(i counter=new HashMap<>(); + for(int n:nums){ + counter.put(n,counter.getOrDefault(n,0)+1); + } + //使用每个数字出现的次数作为排序规则来建立初始化一个优先队列 + PriorityQueue heap=new PriorityQueue<>((n1,n2)-> counter.get(n1)-counter.get(n2)); + //把数字写入优先队列中 + for(int num:counter.keySet()){ + heap.add(num); + //如果优先队列中的元素大于前K个就删除，因为默认是升序。 + if(heap.size()>k){ + heap.poll(); + } + } + //取出前K个元素从优先队列中 + int[] result=new int[k]; + for(int i=0;i> temp=new ArrayList<>(); + int value=0; + int remainder=0; + boolean flag=false; + while (!flag){ + value=a / b; + remainder= a%b; + for (int i=0;i integerIntegerMap = temp.get(i); + //如果相除得到的整数答案和余数在之前出现过，那么就会开始循环。也就是循环节点 + if (integerIntegerMap.containsKey(value) && integerIntegerMap.containsValue(remainder)){ + flag=true; + break; + } + } + HashMap map = new HashMap<>(); + map.put(value,remainder); + temp.add(map); + a=remainder*10; + } + StringBuilder sb=new StringBuilder(); + for (int i=1;i integerIntegerMap = temp.get(i); + integerIntegerMap.forEach((k,v)->{ + sb.append(k); + }); + + } + return sb.toString(); + } + + public static void main(String[] args) { + System.out.println(function(3,7)); + } + +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution5.java b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution5.java new file mode 100644 index 0000000..39e6f26 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution5.java @@ -0,0 +1,43 @@ +package com.algorithm.study.demo.algorithm.leetcode; + +/** + * @author xun2.liu + * @title: Solution5 + * @projectName algorithm-study + * @description: 反转一个单链表。 + * 示例: + * 输入: 1->2->3->4->5->NULL + * 输出: 5->4->3->2->1->NULL + * @date 2020/5/14 20:07 + */ +public class Solution5 { + public static class ListNode { + int val; + ListNode next; + ListNode(int x) { val = x; } + } + public static ListNode reverseList(ListNode head) { + //使用两个指针 + ListNode curr=head; + ListNode prev=null; + while(curr!=null){ + //临时指针。用来存储下一个节点。 + ListNode temp=curr.next; + curr.next=prev; + prev=curr; + curr=temp; + } + return prev; + } + + public static void main(String[] args) { + ListNode a=new ListNode(5); + ListNode b=new ListNode(4); + ListNode c=new ListNode(3); + a.next=b;b.next=c; + ListNode result = reverseList(a); + for (ListNode node=result;node!=null;node=node.next){ + System.out.println(node.val); + } + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution6.java b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution6.java new file mode 100644 index 0000000..ebaa3a5 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution6.java @@ -0,0 +1,49 @@ +package com.algorithm.study.demo.algorithm.leetcode; + +/** + * @author xun2.liu + * @title: Solution6 + * @projectName algorithm-study + * @description: 编写一个程序，找到两个单链表相交的起始节点。 + * 输入：intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3 + * 输出：Reference of the node with value = 8 + * + * 来源：力扣（LeetCode） + * 链接：https://leetcode-cn.com/problems/intersection-of-two-linked-lists + * @date 2020/5/18 19:35 + */ +public class Solution6 { + public static class ListNode { + int val; + ListNode next; + ListNode(int x) { val = x; } + } + public static ListNode getIntersectionNode(ListNode headA, ListNode headB) { + //使用两个指针分别指向headA、headB + //同时遍历两个连表 + //当headA遍历完后指针指向headB，当headB遍历完后指针指向headA + //如此循环当两个指正都为Null的话代表没有相交的节点。 + //如果都两个指针对应的节点相等就返回相等的节点就是相交的节点 + ListNode p1=headA; + ListNode p2=headB; + while(p1!=p2){ + p1=p1==null?headB:p1.next; + p2=p2==null?headA:p2.next; + } + return p1; + } + + public static void main(String[] args) { + ListNode a=new ListNode(5); + ListNode b=new ListNode(4); + a.next=b; + ListNode c=new ListNode(6); + ListNode intersectionNode = getIntersectionNode(a, b); + if (null!=intersectionNode){ + System.out.println(intersectionNode.val); + }else{ + System.out.println("没有相交的节点哦"); + } + + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution7.java b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution7.java new file mode 100644 index 0000000..4ccb886 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution7.java @@ -0,0 +1,76 @@ +package com.algorithm.study.demo.algorithm.leetcode; + +/** + * @author xun2.liu + * @title: Solution7 + * @projectName algorithm-study + * @description: 给定一个链表，返回链表开始入环的第一个节点。 如果链表无环，则返回 null。 + * + * 为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。 + * + * 说明：不允许修改给定的链表。 + * + * 示例 1： + * + * 输入：head = [1,2], pos = 0 + * 输出：tail connects to node index 0 + * 解释：链表中有一个环，其尾部连接到第一个节点。 + * + * 来源：力扣（LeetCode） + * 链接：https://leetcode-cn.com/problems/linked-list-cycle-ii + * + * @date 2020/5/19 17:16 + */ +public class Solution7 { + public static class ListNode { + int val; + ListNode next; + ListNode(int x) { val = x; } + } + + /** + * 快慢指针遍历连表。看是否相遇。如果相遇在判断是否是循环链表。 + * p1=x+y + * p2=x+y+z+y; + * 因为p2是p1的两倍 + * 2*(x+y)=x+y+z+y + * x=z + * @param head + * @return + */ + public static ListNode detectCycle(ListNode head) { + if (null== head || head.next==null){ + return null; + } + //p1指针走一步、p2指针走两步。 + ListNode p1=head; + ListNode p2=head; + while(p2!=null && p2.next!=null){ + p1=p1.next; + p2=p2.next.next; + //如果相等就表示是环形。然后寻找环形入口。 + if(p1==p2){ + //p1指向头结点。找到环形入口 + p1=head; + while(p1!=p2){ + p1=p1.next; + p2=p2.next; + } + return p1; + } + } + return null; + } + public static void main(String[] args) { + ListNode a=new ListNode(5); + ListNode b=new ListNode(4); + ListNode c=new ListNode(6); + ListNode d=new ListNode(-1); + a.next=b; + b.next=c; + c.next=b; +// c.next=b; + ListNode listNode = detectCycle(a); + System.out.println(listNode==null?"":listNode.val); + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution8.java b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution8.java new file mode 100644 index 0000000..2a6a300 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/leetcode/Solution8.java @@ -0,0 +1,68 @@ +package com.algorithm.study.demo.algorithm.leetcode; + +/** + * @author xun2.liu + * @title: Solution8 + * @projectName algorithm-study + * @description: 给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。 + * + * 示例： + * 给定一个链表: 1->2->3->4->5, 和 n = 2. + * 当删除了倒数第二个节点后，链表变为 1->2->3->5. + * 来源：力扣（LeetCode） + * 链接：https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list + * @date 2020/5/20 21:01 + */ +public class Solution8 { + public static class ListNode { + int val; + ListNode next; + ListNode(int x) { val = x; } + } + /** + * 我们可以使用两个指针而不是一个指针。第一个指针从列表的开头向前移动 n+1步，而第二个指针将从列表的开头出发。现在，这两个指针被n个结点分开。 + * 我们通过同时移动两个指针向前来保持这个恒定的间隔，直到第一个指针到达最后一个结点。此时第二个指针将指向从最后一个结点数起的第n个结点。 + * 我们重新链接第二个指针所引用的结点的 next 指针指向该结点的下下个结点。 + * @param head + * @param n + * @return + */ + public static ListNode removeNthFromEnd(ListNode head, int n) { + if (head.next==null){ + return null; + } + //增加一个头部节点，方便删除倒数的节点刚好是第一个节点。 + ListNode temp=new ListNode(-1); + temp.next=head; + ListNode p1=temp; + ListNode p2=temp; + //第一个指针从列表的开头向前移动 n+1步 + for (int i=0;i=0 && r=p1.length()?res:p1; + res=res.length()>=p2.length()?res:p2; + } + return res; + } + public static void main(String[] args) { + System.out.println(longestPalindrome("babad")); + } +} diff --git a/src/main/java/com/algorithm/study/demo/algorithm/lz77/LZ77Codec.java b/src/main/java/com/algorithm/study/demo/algorithm/lz77/LZ77Codec.java new file mode 100644 index 0000000..7c0c777 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/algorithm/lz77/LZ77Codec.java @@ -0,0 +1,124 @@ +package com.algorithm.study.demo.algorithm.lz77; + +/** + * @author xun2.liu + * @title: LZ77Codec + * @projectName algorithm-study + * @description: LZ77算法实现 + * @date 2019/11/25 0:54 + */ +public class LZ77Codec { + /** + * 搜索最大相同字符串的辅助类 + */ + private class SearchResult{ + public int off = 0; + public int length = 0; + } + + /** + * search the max matched string in the slide window; + * 可以使用KMP算法提高效率 + * @param s 需要编码的字符串 + * @param currentPosition 当前位置 + * @return + */ + private SearchResult search(String s,int currentPosition){ + SearchResult result = new SearchResult(); + for (int i = 0; i < currentPosition; i++) { + SearchResult t = new SearchResult(); + for (int j = i; j < currentPosition; j++) { + // 区别已匹配和没有匹配的情况 + if(s.charAt(currentPosition+j-i)==s.charAt(j)){ + // 已经匹配的话 length+1 + // 没有匹配的话 length=1 并计算偏移量 + if(t.length >0){ + t.length++; + }else{ + t.length=1; + // 计算偏移量 + t.off = currentPosition - j; + } + }else { + break; + } + } + if(t.length>result.length){ + result=t; + } + } + return result; + } + /** + * 编码 + * + * 目前发现的问题 + * 1. 从左往右扫描和从右往左扫描off可能不一样 + * @param s 需要编码的字符串 + * @return 已经编码的字符串 + */ + String encoding(String s) { + StringBuilder builder = new StringBuilder(); + // set current coding position pointer to the start of message + int currentPosition = 0; + while (true){ + // search the max matched string in the slide window; + SearchResult result = search(s,currentPosition); + System.out.println("result:"+result.off+" "+result.length+" "+s.substring(currentPosition,currentPosition+result.length+1)); + Character nextChar = s.charAt(currentPosition+result.length); + if(result.length!=0){ + builder.append("(").append(result.off).append(",").append(result.length).append(",").append(nextChar).append(")"); + currentPosition+=result.length+1; + }else { + builder.append("(0,0,").append(nextChar).append(")"); + currentPosition++; + } + if(currentPosition>=s.length()){ + break; + } + } + return builder.toString(); + } + + /** + * 解码 + * @param s 已经编码的字符串 + * @return 已经解码的字符串 + */ + String decoding(String s) { + StringBuilder builder = new StringBuilder(); + // 提取(off,length,next_char) + String[] arr = s.split("\\)\\("); + if (arr.length==0){ + return ""; + } + arr[0]=arr[0].substring(1,arr[0].length()); + if(arr.length>1){ + arr[arr.length-1]=arr[arr.length-1].substring(0,arr[arr.length-1].length()-1); + } + for(String it : arr){ + String[] data = it.split(","); + Integer off = Integer.valueOf(data[0]); + Integer length = Integer.valueOf(data[1]); + String nextChar = data[2]; + Integer iv = builder.length()-off; + for (int i = 0; i < length; i++) { + builder.append(builder.charAt(iv+i)); + } + builder.append(nextChar); + System.out.println("decoding:"+iv+" "+ builder.toString()); + } + return builder.toString(); + } + + public static void main(String[] args) { + LZ77Codec codec = new LZ77Codec(); + String input = "AABCAABCCAABCE"; +// String output = "(0,0,A)(1,1,B)(0,0,C)(4,4,C)(5,4,E)"; + String code = codec.encoding(input); + System.out.println(code); + + String message = codec.decoding(code); + System.out.println(message); + } +} diff --git a/src/main/java/com/algorithm/study/demo/base/AppleMobile.java b/src/main/java/com/algorithm/study/demo/base/AppleMobile.java new file mode 100644 index 0000000..47dd67b --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/base/AppleMobile.java @@ -0,0 +1,5 @@ +package com.algorithm.study.demo.base; + +public abstract class AppleMobile extends Mobile{ + +} diff --git a/src/main/java/com/algorithm/study/demo/base/HuaweiMobile.java b/src/main/java/com/algorithm/study/demo/base/HuaweiMobile.java new file mode 100644 index 0000000..c472308 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/base/HuaweiMobile.java @@ -0,0 +1,13 @@ +package com.algorithm.study.demo.base; + +public class HuaweiMobile extends Mobile{ + + @Override + public void call() { + System.out.println("huawei call"); + } + @Override + public void show(){ + System.out.println("niubi show"); + } +} diff --git a/src/main/java/com/algorithm/study/demo/base/IPerson.java b/src/main/java/com/algorithm/study/demo/base/IPerson.java new file mode 100644 index 0000000..34d7545 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/base/IPerson.java @@ -0,0 +1,5 @@ +package com.algorithm.study.demo.base; + +public interface IPerson { + void eat(); +} diff --git a/src/main/java/com/algorithm/study/demo/base/IphoneXMobile.java b/src/main/java/com/algorithm/study/demo/base/IphoneXMobile.java new file mode 100644 index 0000000..96d35fc --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/base/IphoneXMobile.java @@ -0,0 +1,8 @@ +package com.algorithm.study.demo.base; + +public class IphoneXMobile extends AppleMobile { + @Override + public void call() { + System.out.println("iphonex call"); + } +} diff --git a/src/main/java/com/algorithm/study/demo/base/MainTest.java b/src/main/java/com/algorithm/study/demo/base/MainTest.java new file mode 100644 index 0000000..e5798a7 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/base/MainTest.java @@ -0,0 +1,22 @@ +package com.algorithm.study.demo.base; + +public class MainTest { + private static final String msgg="123"; + public static void main(String[] args) { + MainTest mo=new MainTest(); + Mobile mobile=new IphoneXMobile(); + mobile.call(); + mobile.show(); + + System.out.println(mobile); + Mobile mobile2=new HuaweiMobile(); + mobile2.call(); + mobile2.show(); + + System.out.println(mobile); + + + IPerson per = new YellowPerson(); + per.eat(); + } +} diff --git a/src/main/java/com/algorithm/study/demo/base/Mobile.java b/src/main/java/com/algorithm/study/demo/base/Mobile.java new file mode 100644 index 0000000..e77aebf --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/base/Mobile.java @@ -0,0 +1,9 @@ +package com.algorithm.study.demo.base; + +public abstract class Mobile { + public abstract void call(); + + void show(){ + System.out.println("show"); + } +} diff --git a/src/main/java/com/algorithm/study/demo/base/YellowPerson.java b/src/main/java/com/algorithm/study/demo/base/YellowPerson.java new file mode 100644 index 0000000..4d6945a --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/base/YellowPerson.java @@ -0,0 +1,8 @@ +package com.algorithm.study.demo.base; + +public class YellowPerson implements IPerson { + @Override + public void eat() { + System.out.println("kuaizi eat"); + } +} diff --git a/src/main/java/com/algorithm/study/demo/datastructure/graph/Mgraph.java b/src/main/java/com/algorithm/study/demo/datastructure/graph/Mgraph.java new file mode 100644 index 0000000..f74335e --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/datastructure/graph/Mgraph.java @@ -0,0 +1,120 @@ +package com.algorithm.study.demo.datastructure.graph; + +import java.util.LinkedList; +import java.util.Queue; + +/** + * 图 + * @Author: liuxun + * @CreateDate: 2019/2/21 上午10:40 + * @Version: 1.0 + */ +public class Mgraph { + private int v;//顶点的个数 + private LinkedList adj[];//邻接表 + + public Mgraph(int capaCity){ + v=capaCity; + adj=new LinkedList[capaCity]; + for (int i=0;i(); + } + } + + /** + * 添加数据 + * @param oSide + * @param rSide + */ + private void add(int oSide,int rSide){ + adj[oSide].add(rSide); + adj[rSide].add(oSide); + } + + /** + * 构造一个无向图 + */ + private void createGraph() { + // 0 -- 1 -- 2 + // | | | + // 3 -- 4 -- 5 + // | | | + // 6 -- 7 -- 8 + add(0,1);//add(1,0); + add(0,3);//add(3,0); + + add(1,2);//add(2,1); + add(1,4);// add(4,1); + + add(2,5);//add(5,2); + + add(3,4);//add(4,3); + add(3,6);//add(6,3); + + add(4,5);//add(5,4); + add(4,7);// add(7,4); + + add(5,8);//add(8,5); + + add(6,7);//add(7,6); + + add(7,8);//add(8,7); + } + private void print(int[] prev, int oSide, int rSide) { + if (prev[rSide] != -1 && oSide != rSide) { + print(prev, oSide, prev[rSide]); + } + System.out.print(rSide + " "); + } + + /** + * 广度优先搜索，从字面意思理解，它就是一种“地毯式”的搜索策略，先查找离起始顶点最近的，然后是次近的，依次往外搜索，层层递进。 + * + * 在这里三个重要的核心辅助变量 visited、queue、prev。 + * visited 记录已经被访问的顶点，避免顶点被重复访问 + * queue 用来存储已经被访问、但相连的顶点还没有被访问的顶点的这样的一个队列。 + * prev 记录搜索路径，它是反向存储，便于后续正向打印输出图的路径。 + * @param oSide + * @param rSide + */ + private void bfs(int oSide, int rSide) { + if (oSide == rSide) return; + + boolean[] visited = new boolean[v]; + visited[oSide] = true; + Queue queue = new LinkedList<>(); + queue.offer(oSide); + int[] prev = new int[v]; + for (int i = 0; i < v; i++) { + prev[i] = -1; + } + while (!queue.isEmpty()) { + int index = queue.poll(); + for (int j = 0; j < adj[index].size(); j++) { + int value = adj[index].get(j); + if (!visited[value]) { + prev[value] = index; + if (value == rSide) { + print(prev, oSide, rSide); + } + visited[value] = true; + queue.offer(value); + } + } + } + } + + public static void main(String[] args) { + int count = 9; + Mgraph graph = new Mgraph(count); + // 0 -- 1 -- 2 + // | | | + // 3 -- 4 -- 5 + // | | | + // 6 -- 7 -- 8 + graph.createGraph(); + System.out.println("BFS(广度优先搜索)"); + graph.bfs(0,6); + + } +} diff --git a/src/main/java/com/algorithm/study/demo/datastructure/heap/Heap.java b/src/main/java/com/algorithm/study/demo/datastructure/heap/Heap.java new file mode 100644 index 0000000..8870957 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/datastructure/heap/Heap.java @@ -0,0 +1,106 @@ +package com.algorithm.study.demo.datastructure.heap; + +import com.alibaba.fastjson.JSON; + +/** + * 堆 + * 堆就是一颗完全二叉树 + * 堆中每一个节点的值都必须大于等于（或者小于等于）其子树（左右子节点）中每个节点的值。 + * 堆顶存储的就是最大值或者最小值 + * 用数组存储非常节省存储空间，下标为id的节点的左子节点就是i*2的节点，右子节点的下标为i*2+1。父节点就是i/2的节点。 + * + * @Author: liuxun + * @CreateDate: 2019/1/31 上午10:30 + * @Version: 1.0 + */ +public class Heap { + private int[] data; + private int count;//堆中元素的个数 + private int capacity;//堆容量 + + /** + * 初始化堆 + * @param capacity + */ + public Heap(int capacity){ + this.capacity=capacity; + data=new int[capacity]; + count=0; + } + + /** + * 往堆中插入数据 + * 从下往上的方法堆化 + * @param d + */ + public void insert(int d){ + if (count>=capacity) return;//堆满了 + count++;//从下标为1开始填充数据 + data[count]=d; + int i=count; + while (i/2>0 && data[i]>data[i/2]){ + //交换数据 + swap(data,i,i/2); + i=i/2;//切到父节点 + } + } + + private static void swap(int[] a,int i,int n) { + int temp=a[i]; + a[i]=a[n]; + a[n]=temp; + } + + /** + * 删除堆顶元素 + * 把最后一个元素放到对顶，然后从上到下的方法堆化。 + */ + public void removeMax(){ + data[1]=data[count]; + count--; + int i=1; + heapify(data,count,i); + } + + private static void buildHeap(int[] a, int n) { + for (int i = n/2; i >= 1; --i) { + heapify(a, n, i); + } + System.out.println(JSON.toJSONString(a)); + } + + private static void heapify(int[] a, int n, int i) { + while (true) { + int maxPos = i; + if (i*2 <= n && a[i] < a[i*2]) maxPos = i*2; + if (i*2+1 <= n && a[maxPos] < a[i*2+1]) maxPos = i*2+1; + if (maxPos == i) break; + swap(a, i, maxPos); + i = maxPos; + } + } + + // n 表示数据的个数，数组 a 中的数据从下标 1 到 n 的位置。 + public static void sort(int[] a, int n) { + buildHeap(a, n);//构建堆 + int k = n; + while (k > 1) { + swap(a, 1, k);//堆顶跟最后一个元素交换 + --k;//每次k-1，也就去取出 + heapify(a, k, 1);//1-k堆化 + } + System.out.println(JSON.toJSONString(a)); + } + + + public static void main(String[] args) { +// Heap heap=new Heap(6); +// for (int i=1;i<=5;i++){ +// heap.insert(i); +// } +// System.out.println(heap.count); + + int[] d = new int[]{0, 5, 2, 1, 4, 3}; + sort(d,d.length-1); + } +} diff --git a/src/main/java/com/algorithm/study/demo/datastructure/heap/MidNumCount.java b/src/main/java/com/algorithm/study/demo/datastructure/heap/MidNumCount.java new file mode 100644 index 0000000..28224f4 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/datastructure/heap/MidNumCount.java @@ -0,0 +1,83 @@ +package com.algorithm.study.demo.datastructure.heap; + +import java.util.Comparator; +import java.util.PriorityQueue; + +/** + * 求中位数的问题 + * @Author: liuxun + * @CreateDate: 2019/1/31 下午1:57 + * @Version: 1.0 + */ +public class MidNumCount { + + /** 大顶堆，用来存储前半部分的数据，如果完整为100，那此存储的为0-50 */ + private PriorityQueue firstBigHeap =new PriorityQueue<>(11, new Comparator() { + @Override + public int compare(Integer o1, Integer o2) { + return o2-o1; + } + }); + + /** 小顶堆,用来存储后半部分的数据，如果完整为100,那此存储的为51-100 */ + private PriorityQueue afterLittleHeap = new PriorityQueue<>(11); + + /** 元素的个数 */ + private int count; + + /** + * 插入数据 + * + * @param num 当前动态的数据集 + */ + public void putNum(int num) { + + count++; + + // 1,如果堆为空，则插入大顶堆中 + if (firstBigHeap.isEmpty() && afterLittleHeap.isEmpty()) { + firstBigHeap.offer(num); + return; + } + //奇数 + if (((firstBigHeap.size()+afterLittleHeap.size()) & 1)==1){ + //如果数据当前元素比大顶堆中的元素大，则插入小顶堆中 + if (afterLittleHeap.peek() > num) { + afterLittleHeap.offer(num); + + } + }else{ + + } + + } + + /** + * 返回中位数的数据 + * + * @return + */ + public int getMidValue() { + return firstBigHeap.peek(); + } + + /** + * 从一个堆向另一个堆中移动元素 + * + * @param src + * @param out + */ + private void move(PriorityQueue src, PriorityQueue out, int runNum) { + for (int i = 0; i < runNum; i++) { + out.offer(src.poll()); + } + } + + public static void main(String[] args) { + + int[] da=new int[]{5,6,4,1,2}; + + System.out.println(6>>1); + System.out.println(6&1); + } +} diff --git a/src/main/java/com/algorithm/study/demo/datastructure/heap/TopkCount.java b/src/main/java/com/algorithm/study/demo/datastructure/heap/TopkCount.java new file mode 100644 index 0000000..54b1cdc --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/datastructure/heap/TopkCount.java @@ -0,0 +1,100 @@ +package com.algorithm.study.demo.datastructure.heap; + +import com.alibaba.fastjson.JSON; + +import java.util.PriorityQueue; + +/** + * @Author: liuxun + * @CreateDate: 2019/1/30 下午8:10 + * @Version: 1.0 + */ +public class TopkCount { + /** + * 求数据中前K大数据 + * + * @param data + * @param k + * @return + */ + public static int topk(int[] data, int k) { + PriorityQueue queue = new PriorityQueue<>(k); + + for (int i = 0; i < data.length; i++) { + if (queue.size() < k) { + queue.offer(data[i]); + } else { + int value = queue.peek(); + // 如果发现数据比堆顶元素大，则加入到小顶堆中 + if (data[i] > value) { + queue.poll();//删除 + queue.offer(data[i]); + } + } + } + return queue.peek(); + } + + /** + * 手动实现堆然后查找topK + * @param nums + * @param k + * @return + */ + public static int findKthLargest(int[] nums, int k) { + //构建一个小顶堆数组 + int[] out = new int[k]; + //小顶堆元素数量 + int count = 0; + //循环处理数组 + for (int i=0;iout[0]){ + out[0] = nums[i]; + up2down(out,0); + } + } + } + return out[0]; + } + + //从下向上调整,构建小顶堆 + public static void down2up(int[] nums,int i){ + int parent = (i-1)/2; + while(parent>=0&&nums[i] { // 定义一个内部类Node，代表链表的节点 - private class Node { - private E data;// 保存数据 + public class Node { + private E value;// 保存数据 private Node next;// 指向下个节点的引用 // 无参构造器 public Node() { } + // 初始化全部属性的构造器 public Node(E data, Node next) { - this.data = data; + this.value = data; this.next = next; } } + private Node data;// 保存头结点 - private int size=0;// 保存已含有的节点数 + private int size = 0;// 保存已含有的节点数 - public MLinkList(){ - this.data=null;//初始化一个空的头结点 + public MLinkList() { + this.data = null;//初始化一个空的头结点 } /** * 添加一个头结点 + * * @param element */ - public void addFirst(E element){ - if (data==null){ - data=new Node(element,null); - }else{ - Node temp=new Node(element,null); - temp.next=data; - data=temp; + public void addFirst(E element) { + if (data == null) { + data = new Node(element, null); + } else { + Node temp = new Node(element, null); + temp.next = data; + data = temp; } size++; } + /** * 删除一个头结点 + * * @return */ - public E deleteFirst(){ - Node temp=data; - data=temp.next; - temp.next=null;//释放应用 - size--; - return temp.data; + public E deleteFirst() { + Node current = data; + E val = current.value; + current.value = current.next.value; + current.next = current.next.next; + return val; } /** * 在index插入节点 + * * @param index * @param element */ - public void add(int index,E element){ + public void add(int index, E element) { checkPositionIndex(index); - if (index==0){ + if (index == 0) { addFirst(element); return; } - Node newNode=new Node(element,null);//新的结点 - Node current=data;//保存index当前的结点 - int i=1;//默认是第i个结点 - while (isize-1){ - throw new IndexOutOfBoundsException("数组越界Index: "+index+", Size: "+size); + if (index < 0 || index > size - 1) { + throw new IndexOutOfBoundsException("数组越界Index: " + index + ", Size: " + size); } } - public int size(){ + + public int size() { return size; } @Override - public String toString(){ - StringBuilder sb=new StringBuilder(); - int j=1; - Node temp=data; - while (j mLinkList = new MLinkList(); + mLinkList.add(4); + mLinkList.add(1); + mLinkList.add(8); + mLinkList.add(4); + mLinkList.add(5); +// System.out.println(mLinkList.toString()); +// mLinkList.delete("b"); +// mLinkList.delete(0); +// System.out.println(mLinkList.size); +// mLinkList.reverseLinkedList(); System.out.println(mLinkList.toString()); } diff --git a/src/main/java/com/algorithm/study/demo/datastructure/linear/Solution.java b/src/main/java/com/algorithm/study/demo/datastructure/linear/Solution.java new file mode 100644 index 0000000..2e0f315 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/datastructure/linear/Solution.java @@ -0,0 +1,142 @@ +package com.algorithm.study.demo.datastructure.linear; + +/** + * 单链表链表算法题 + * @Author: liuxun + * @CreateDate: 2019/2/12 下午3:43 + * @Version: 1.0 + */ +public class Solution { + + /** + * 环形链表检测 + * @param head + * @return + */ + public static ListNode detectCycle(ListNode head) { + ListNode temp=head; + ListNode temp2=head; + while(temp!=null && temp.next.next!=null){ + //通过快慢指针遍历 + temp2=temp2.next.next; + temp=temp.next; + if (temp==temp2){ + //寻找环形入口 + ListNode q = head; + while(temp2!=q){ + temp2=temp2.next; + q=q.next; + } + return q; + } + } + return null; + } + + /*** + * 单链表的前K个的逆序输出 + * + */ + public static void reversedTopK(ListNode head,int k){ + if (head==null || head.next==null){ + return; + } + printNode(head); + int count=1; + ListNode p1=head; + ListNode p2=head.next; + ListNode p3=null; + while (count"); + } + } + + public static void main(String[] args) { + ListNode head=new ListNode(3); + ListNode head0=new ListNode(1); + + ListNode head1=new ListNode(2); + ListNode head2=new ListNode(-4); + +// head.next=head0; +// head0.next=head1; +// head1.next=head2; +// head2.next=head1; +// ListNode c=Solution.detectCycle(head); +// System.out.println(c==null?-1:1); + + head.next=head0; + head0.next=head1; + head1.next=head2; + head2.next=null; +// Solution.reverseLinkedList(head); +// Solution.reversedTopK(head,2); + + Solution.findKthToTail(head,2); + } +} diff --git a/src/main/java/com/algorithm/study/demo/datastructure/tree/AVLBinTree.java b/src/main/java/com/algorithm/study/demo/datastructure/tree/AVLBinTree.java deleted file mode 100644 index 87e47a4..0000000 --- a/src/main/java/com/algorithm/study/demo/datastructure/tree/AVLBinTree.java +++ /dev/null @@ -1,10 +0,0 @@ -package com.algorithm.study.demo.datastructure.tree; - -/** - * AVL二叉搜索 - * @Author: liuxun - * @CreateDate: 2018/10/22 上午11:26 - * @Version: 1.0 - */ -public class AVLBinTree { -} diff --git a/src/main/java/com/algorithm/study/demo/datastructure/tree/LinkBinTree.java b/src/main/java/com/algorithm/study/demo/datastructure/tree/LinkBinTree.java index 805abcc..ce3c9e4 100644 --- a/src/main/java/com/algorithm/study/demo/datastructure/tree/LinkBinTree.java +++ b/src/main/java/com/algorithm/study/demo/datastructure/tree/LinkBinTree.java @@ -1,29 +1,32 @@ package com.algorithm.study.demo.datastructure.tree; -import java.util.ArrayList; -import java.util.LinkedList; -import java.util.Queue; -import java.util.Stack; +import com.alibaba.fastjson.JSON; +import org.testng.collections.Lists; + +import java.util.*; /** - * * 二叉搜索树链表存储 - 前序遍历：根节点->左子树->右子树 - 中序遍历：左子树->根节点->右子树 - 后序遍历：左子树->右子树->根节点 + * 前序遍历：根节点->左子树->右子树 + * 中序遍历：左子树->根节点->右子树 + * 后序遍历：左子树->右子树->根节点 * Created by liuxun on 2017/6/29. */ public class LinkBinTree { - public static class TreeNode{ + public static class TreeNode { Integer data;//节点数据 TreeNode left;//左子节点数据 TreeNode right;//右子节点数据 - TreeNode(){} - TreeNode(Integer data){ + + TreeNode() { + } + + TreeNode(Integer data) { this.data = data; this.left = null; this.right = null; } + public TreeNode(Integer data, TreeNode left, TreeNode right) { this.data = data; this.left = left; @@ -32,160 +35,194 @@ public TreeNode(Integer data, TreeNode left, TreeNode right) { } private TreeNode root; - /**初始化空的二叉树**/ - public LinkBinTree(){ - root=new TreeNode(); + + /** + * 初始化空的二叉树 + **/ + public LinkBinTree() { + root = new TreeNode(); } - /**指定一个默认的根二叉树**/ - public LinkBinTree(Integer d){ - root=new TreeNode(d); + + /** + * 指定一个默认的根二叉树 + **/ + public LinkBinTree(Integer d) { + root = new TreeNode(d); } - /**判断二叉树是否为空**/ - public boolean isEmpty(){ - return root.data==null; + + /** + * 判断二叉树是否为空 + **/ + public boolean isEmpty() { + return root.data == null; } - /**获取根节点**/ - public TreeNode getRoot(){ - if (isEmpty()){ + + /** + * 获取根节点 + **/ + public TreeNode getRoot() { + if (isEmpty()) { throw new RuntimeException("树为空，无法获取根节点"); } return root; } - /**获取树的深度**/ - public int getDeep(TreeNode t){ - if (t==null){ + /** + * 获取树的深度 + **/ + public int getDeep(TreeNode t) { + if (t == null) { return 0; } - int l=getDeep(t.left); - int r=getDeep(t.right); - return l>r?(l+1):(r+1); + int l = getDeep(t.left); + int r = getDeep(t.right); + return l > r ? (l + 1) : (r + 1); } - /**获取树的最小深度**/ - public int getMinDeep(TreeNode t){ - if (t==null){ + + /** + * 获取树的最小深度 + **/ + public int getMinDeep(TreeNode t) { + if (t == null) { return 0; } - if (t.left==null && t.right==null){ + if (t.left == null && t.right == null) { return 1; } - int l=getMinDeep(t.left); - int r=getMinDeep(t.right); - return lt.data){ - if(t.right!=null){ - add(t.right,value); - } - else{ + + private void add(TreeNode t, int value) { + if (value > t.data) { + if (t.right != null) { + add(t.right, value); + } else { t.right = new TreeNode(value); } - } - else{ - if(t.left!=null){ - add(t.left,value); - } - else{ + } else { + if (t.left != null) { + add(t.left, value); + } else { t.left = new TreeNode(value); } } } - private void add2(TreeNode t,int value){ - TreeNode node=new TreeNode(value); - TreeNode current=t; - while(current!=null){ - TreeNode parentNode=current; - if (current.data>value){ - current=current.left; - if (current==null){ - parentNode.left=node; + + private void add2(TreeNode t, int value) { + if (null == t.data) { + t.data = value; + return; + } + TreeNode node = new TreeNode(value); + TreeNode current = t; + while (true) { + TreeNode parentNode = current; + if (current.data > value) { + current = current.left; + if (current == null) { + parentNode.left = node; return; } - }else{ - current=current.right; - if (current==null){ - parentNode.right=node; + } else { + current = current.right; + if (current == null) { + parentNode.right = node; return; } } } } + /** * 递归从根节点开始插入数据，大于根节点放在右子树，小于根节点放在左子树 + * * @param value */ - public void add(int value){ - add(root,value); + public void add(int value) { + add(root, value); } + /** * 非递归模式插入数据 * 从根节点开始插入数据，大于根节点放在右子树，小于根节点放在左子树 + * * @param value */ - public void add2(int value){ - add2(root,value); + public void add2(int value) { + add2(root, value); } + /** * 前序遍历 * 如果树为空返回，如果不为空首先从根节点开始遍历，然后先前序遍历左子树，最后前序遍历右子树。 */ - public void preOrderTraverse(TreeNode t){ - if (t==null) { + public void preOrderTraverse(TreeNode t) { + if (t == null) { return; } System.out.println(t.data); preOrderTraverse(t.left); preOrderTraverse(t.right); } - public void preOrderTraverse(){ + + public void preOrderTraverse() { preOrderTraverse(root); } /** * 非递归前序遍历 + * * @param t */ public void preOrderTraverse2(TreeNode t) { - if (t==null) { + if (t == null) { return; } - Stack stack=new Stack<>(); - while(t!=null || !stack.isEmpty()){ - while (t!=null){ + Stack stack = new Stack<>(); + while (t != null || !stack.isEmpty()) { + while (t != null) { System.out.println(t.data); stack.push(t); - t=t.left; + t = t.left; } - if (!stack.isEmpty()){ - t=stack.pop(); - t=t.right; + if (!stack.isEmpty()) { + t = stack.pop(); + t = t.right; } } } - public void preOrderTraverse2(){ + + public void preOrderTraverse2() { preOrderTraverse2(root); } @@ -193,64 +230,71 @@ public void preOrderTraverse2(){ * 中序遍历 * 如果树为空返回，从根节点开始，中序遍历左子树，然后访问根节点，最后中序遍历右子树。 */ - public void inOrderTraverse(TreeNode t){ - if (t==null) { + public void inOrderTraverse(TreeNode t) { + if (t == null) { return; } inOrderTraverse(t.left); System.out.println(t.data); inOrderTraverse(t.right); } - public void inOrderTraverse(){ + + public void inOrderTraverse() { inOrderTraverse(root); } /** * 非递归中序遍历 + * * @param t */ - public void inOrderTraverse2(TreeNode t){ - if (t==null) { + public void inOrderTraverse2(TreeNode t) { + if (t == null) { return; } - Stack stack=new Stack<>(); - while (t!=null || !stack.isEmpty()){ - while (t!=null){ + Stack stack = new Stack<>(); + while (t != null || !stack.isEmpty()) { + while (t != null) { stack.push(t); - t=t.left; + t = t.left; } - if (!stack.isEmpty()){ - t=stack.pop(); + if (!stack.isEmpty()) { + t = stack.pop(); System.out.println(t.data); - t=t.right; + t = t.right; } } } - public void inOrderTraverse2(){ + + public void inOrderTraverse2() { inOrderTraverse2(root); } + /** * 后续遍历 + * * @param t */ - public void postOrderTraverse(TreeNode t){ - if (t==null) { + public void postOrderTraverse(TreeNode t) { + if (t == null) { return; } postOrderTraverse(t.left); postOrderTraverse(t.right); System.out.println(t.data); } - public void postOrderTraverse(){ + + public void postOrderTraverse() { postOrderTraverse(root); } /** * 非递归后续遍历 + * * @param root */ - public void postOrderTraverse2(TreeNode root){ + public void postOrderTraverse2(TreeNode root) { Stack s = new Stack(); Stack s2 = new Stack(); Integer i = new Integer(1); //0表示对应位置上的节点还没有遍历过右节点，1表示已经遍历过 @@ -273,63 +317,199 @@ public void postOrderTraverse2(TreeNode root){ } } } - public void postOrderTraverse2(){ + + public void postOrderTraverse2() { postOrderTraverse2(root); } + /** * 层级遍历 + * * @param t */ - public void divOrderTraverse(TreeNode t){ - if (t==null) { - return; + public List> divOrderTraverse(TreeNode t) { + if (t == null) { + return new ArrayList>(); } - Queue queue = new LinkedList() ; + //初始化队列只包含一个节点 root 和层次编号 0 ： level = 0。 + List> levels = new ArrayList<>(); + Queue queue = new LinkedList(); queue.add(root); - while(queue.size() != 0) - { + //树的层数 + int level = 0; + while (queue.size() != 0) { + //插入一个空列表，开始当前层的算法。 + levels.add(new ArrayList<>()); int len = queue.size(); - for(int i=0;i > levelOrder(TreeNode root) { + if (root == null) { + return new ArrayList>(); + } + List> lists = new ArrayList<>(); + Queue queue = new LinkedList(); + queue.offer(root); + while (queue.size() > 0) { + LinkedList levelList = new LinkedList<>(); + for (int i = queue.size(); i > 0; i--) { + TreeNode node = queue.poll(); + if ((lists.size() & 1) == 1) { + //奇数层放到队列尾部 + levelList.addLast(node.data); + } else { + //偶数层放到队列头部 + levelList.addFirst(node.data); + } + if (node.right != null) { + queue.offer(node.right); + } + if (node.left != null) { + queue.offer(node.left); + } + } + lists.add(levelList); + } + return lists; + } + + /** + * 层级遍历1 + */ + public void divOrderTraverse() { + List> lists = divOrderTraverse(root); + System.out.println(JSON.toJSONString(lists)); + } + + /** + * 层级遍历2 + */ + public void levelOrder() { + List> lists = levelOrder(root); + System.out.println(JSON.toJSONString(lists)); + } + + /** + * 序列化树 + * + * @param root + * @return + */ + public String serialize(TreeNode root) { + if (root == null) { + return null; + } + //使用层序遍历 + Queue que = new LinkedList(); + StringBuilder sb = new StringBuilder("["); + que.add(root); + while (que.size() > 0) { + int currSize = que.size(); + for (int i = 0; i < currSize; i++) { + TreeNode node = que.poll(); + if (node != null) { + sb.append(node.data).append(","); + que.add(node.left); + que.add(node.right); + } else { + sb.append("null,"); + } + } + } + return sb.deleteCharAt(sb.length() - 1).append("]").toString(); } - /**区间搜索**/ - private void searchSection(TreeNode t,int k1,int k2,ArrayList result){ - if (t==null){ + + public String serialize() { + String serialize = serialize(root); + System.out.println(serialize); + return serialize; + } + + // Decodes your encoded data to tree. + public TreeNode deserialize(String data) { + if (data == null || data.length() == 0) { + return null; + } + data = data.substring(1, data.length() - 1); + String[] arrData = data.split(","); + //填充根节点 + TreeNode tree = new TreeNode(Integer.valueOf(arrData[0])); + Queue que = new LinkedList<>(); + que.add(tree); + int i = 1; + while (que.size() > 0 && i < arrData.length) { + TreeNode currNode = que.poll(); + + if (arrData[i].equals("null")) { + currNode.left = null; + } else { + TreeNode treeLeft = new TreeNode(Integer.valueOf(arrData[i])); + currNode.left = treeLeft; + que.add(treeLeft); + } + i++; + + if (arrData[i].equals("null")) { + currNode.right = null; + } else { + TreeNode treeRight = new TreeNode(Integer.valueOf(arrData[i])); + currNode.right = treeRight; + que.add(treeRight); + } + i++; + } + return tree; + } + + /** + * 区间搜索 + **/ + private void searchSection(TreeNode t, int k1, int k2, ArrayList result) { + if (t == null) { return; } - if(t.data>k1){ - searchSection(t.left,k1,k2,result); + if (t.data > k1) { + searchSection(t.left, k1, k2, result); } - if(t.data>=k1&&t.data<=k2){ + if (t.data >= k1 && t.data <= k2) { result.add(t.data); } - if(t.datakey){ - currnode=currnode.left; - }else if (currnode.data key) { + currnode = currnode.left; + } else if (currnode.data < key) { + currnode = currnode.right; + } else { return currnode; } } @@ -338,39 +518,42 @@ public TreeNode find(int key){ /** * 查找最小值 + * * @return */ - public TreeNode findMin(){ - TreeNode current=root; - TreeNode minNode=current; - while(current!=null){ - minNode=current; - current=current.left; + public TreeNode findMin() { + TreeNode current = root; + TreeNode minNode = current; + while (current != null) { + minNode = current; + current = current.left; } return minNode; } /** * 查找最大值 + * * @return */ - public TreeNode findMax(){ - TreeNode current=root; - TreeNode maxNode=current; - while(current!=null){ - maxNode=current; - current=current.right; + public TreeNode findMax() { + TreeNode current = root; + TreeNode maxNode = current; + while (current != null) { + maxNode = current; + current = current.right; } return maxNode; } + public static void main(String[] args) { - int[] ls=new int[]{30,9,8}; - LinkBinTree linkBinTree=new LinkBinTree(ls[0]); - for (int i=1;i list=new ArrayList(); // linkBinTree.searchSection(linkBinTree.getRoot(),10,20,list); // System.out.println("区间查询"+list.toString()); -// System.out.println("-------------递归遍历----------------"); + System.out.println("-------------递归遍历----------------"); // linkBinTree.preOrderTraverse();//前序遍历 从根节点开始遍历 -// System.out.println("-----------------------------"); + System.out.println("-----------------------------"); // linkBinTree.inOrderTraverse();//中序遍历 从根节点开始 -// System.out.println("-----------------------------"); + System.out.println("-----------------------------"); // linkBinTree.postOrderTraverse();//后序遍历 -// System.out.println("-----------------------------"); + System.out.println("-----------------------------"); // linkBinTree.divOrderTraverse();//层次遍历 +// linkBinTree.levelOrder(); + //序列化、反序列化树 + System.out.println("-----------------------------"); + TreeNode deserializeTree = linkBinTree.deserialize(linkBinTree.serialize()); + linkBinTree.levelOrder(deserializeTree); // //前序遍历：根节点->左子树->右子树 // //中序遍历：左子树->根节点->右子树 // //后序遍历：左子树->右子树->根节点 // System.out.println(); // System.out.println("-------------非递归遍历----------------"); - linkBinTree.preOrderTraverse2();//前序遍历 +// linkBinTree.preOrderTraverse2();//前序遍历 // System.out.println("-----------------------------"); // linkBinTree.inOrderTraverse2();//中序遍历 // System.out.println("-----------------------------"); // linkBinTree.postOrderTraverse2();//后序遍历 - //二叉查找树搜索 - TreeNode node = linkBinTree.find(9); - System.out.println(node.data); - System.out.println("最小值为："+linkBinTree.findMin().data); + //二叉查找树搜索 +// TreeNode node = linkBinTree.find(9); +// System.out.println(node.data); +// System.out.println("最小值为："+linkBinTree.findMin().data); + } } diff --git a/src/main/java/com/algorithm/study/demo/enums/Calculator.java b/src/main/java/com/algorithm/study/demo/enums/Calculator.java new file mode 100644 index 0000000..2e4a0c2 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/enums/Calculator.java @@ -0,0 +1,15 @@ +package com.algorithm.study.demo.enums; + +/** + * @author xun2.liu + * @title: Calculator + * @projectName algorithm-study + * @description: TODO + * @date 2020/1/7 14:51 + */ +public class Calculator{ + + public int apply(int a, int b,Operator operator) { + return operator.apply(a,b); + } +} diff --git a/src/main/java/com/algorithm/study/demo/enums/MainTest.java b/src/main/java/com/algorithm/study/demo/enums/MainTest.java new file mode 100644 index 0000000..85a7c04 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/enums/MainTest.java @@ -0,0 +1,19 @@ +package com.algorithm.study.demo.enums; + +import java.text.ParseException; + +/** + * @author xun2.liu + * @title: MainTest + * @projectName algorithm-study + * @description: 枚举类解决IF ESLE问题 + * @date 2019/12/13 16:32 + */ +public class MainTest{ + public static void main(String[] args) throws ParseException { + Calculator calculator=new Calculator(); + int result=calculator.apply(2,4,Operator.ADD); + System.out.println(result); + } + +} diff --git a/src/main/java/com/algorithm/study/demo/enums/Operator.java b/src/main/java/com/algorithm/study/demo/enums/Operator.java new file mode 100644 index 0000000..54201b9 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/enums/Operator.java @@ -0,0 +1,48 @@ +package com.algorithm.study.demo.enums; + +/** + * @author xun2.liu + * @title: Operator + * @projectName algorithm-study + * @description: TODO + * @date 2019/12/13 16:31 + */ +public enum Operator { + + ADD { + @Override + public int apply(int a, int b) { + return a + b; + } + }, + + MULTIPLY { + @Override + public int apply(int a, int b) { + return a * b; + } + }, + + SUBTRACT { + @Override + public int apply(int a, int b) { + return a - b; + } + }, + + DIVIDE { + @Override + public int apply(int a, int b) { + return a / b; + } + }, + + MODULO { + @Override + public int apply(int a, int b) { + return a % b; + } + }; + + public abstract int apply(int a, int b); +} \ No newline at end of file diff --git a/src/main/java/com/algorithm/study/demo/guava/MainTest.java b/src/main/java/com/algorithm/study/demo/guava/MainTest.java new file mode 100644 index 0000000..445274e --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/guava/MainTest.java @@ -0,0 +1,40 @@ +//package com.algorithm.study.demo.guava; +// +//import com.github.rholder.retry.*; +//import com.google.common.base.Predicates; +// +//import java.io.*; +//import java.text.SimpleDateFormat; +//import java.time.LocalDateTime; +//import java.time.format.DateTimeFormatter; +//import java.util.concurrent.ExecutionException; +//import java.util.concurrent.TimeUnit; +// +///** +// * guava 重试机制 +// * @Author: liuxun +// * @CreateDate: 2019/1/2 上午11:29 +// * @Version: 1.0 +// */ +//public class MainTest { +// public static void main(String[] args) { +// //定义重试机制 +// Retryer retryer = RetryerBuilder.newBuilder() +// .retryIfException() //设置异常重试 +// .retryIfResult(Predicates.equalTo(true)) //call方法返回true重试 +// .withWaitStrategy(WaitStrategies.fixedWait(10, TimeUnit.SECONDS)) //设置10秒后重试 +// .withStopStrategy(StopStrategies.stopAfterAttempt(3)).build(); //设置重试次数 超过将出异常 +// try { +// retryer.call(() -> { +// //这里写你的业务逻辑代码 +// System.out.println("11111111111111111122222"); +// return true; //需要重试返回true +// }); +// } catch (ExecutionException e) { +// e.printStackTrace(); +// } catch (RetryException e) { +// e.printStackTrace(); +// } +// } +//} +// diff --git a/src/main/java/com/algorithm/study/demo/java8/FunctionTest.java b/src/main/java/com/algorithm/study/demo/java8/FunctionTest.java new file mode 100644 index 0000000..1706fae --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/java8/FunctionTest.java @@ -0,0 +1,95 @@ +package com.algorithm.study.demo.java8; + +import org.testng.Assert; + +import java.util.function.Consumer; +import java.util.function.Function; +import java.util.function.Predicate; + +/** + * @author xun2.liu + * @title: FunctionTest + * @projectName algorithm-study + * @description: Consumer实例 + * @date 2019/12/20 15:19 + */ +public class FunctionTest { + public static void main(String[] args) { +// consumerTest(); + functionTest(); + } + + /*** + * Consumer是一个函数式编程接口； 顾名思义，Consumer的意思就是消费，即针对某个东西我们来使用它，因此它包含有一个有输入而无输出的accept接口方法； + * 除accept方法，它还包含有andThen这个方法； + */ + public static void consumerTest() { + Consumer f = System.out::println; + Consumer f2 = n -> System.out.println(n + "-F2"); + + //执行完F后再执行F2的Accept方法 + f.andThen(f2).accept("test"); + + //连续执行F的Accept方法 +// f.andThen(f).andThen(f).andThen(f).accept("test1"); + } + + /** + * Function测试 + * Function也是一个函数式编程接口；它代表的含义是“函数”，而函数经常是有输入输出的，因此它含有一个apply方法， + * 包含一个输入与一个输出；除apply方法外，它还有compose与andThen及indentity三个方法 + */ + public static void functionTest() { + Function f = s -> s++; + Function g = s -> s * 2; + + /** + * 下面表示在执行F时，先执行G，并且执行F时使用G的输出当作输入。 + * 相当于以下代码： + * Integer a = g.apply(1); + * System.out.println(f.apply(a)); + */ + System.out.println(f.compose(g).apply(1)); + + /** + * 表示执行F的Apply后使用其返回的值当作输入再执行G的Apply； + * 相当于以下代码 + * Integer a = f.apply(1); + * System.out.println(g.apply(a)); + */ + System.out.println(f.andThen(g).apply(1)); + + /** + * identity方法会返回一个不进行任何处理的Function，即输出与输入值相等； + */ + System.out.println(Function.identity().apply("a")); + } + + + + /** + * Predicate测试 + * Predicate为函数式接口，predicate的中文意思是“断定”，即判断的意思，判断某个东西是否满足某种条件； + * 因此它包含test方法，根据输入值来做逻辑判断，其结果为True或者False。 + */ + private static void predicateTest() { + Predicate p = o -> o.equals("test"); + Predicate g = o -> o.startsWith("t"); + + /** + * negate: 用于对原来的Predicate做取反处理； + * 如当调用p.test("test")为True时，调用p.negate().test("test")就会是False； + */ + Assert.assertFalse(p.negate().test("test")); + + /** + * and: 针对同一输入值，多个Predicate均返回True时返回True，否则返回False； + */ + Assert.assertTrue(p.and(g).test("test")); + + /** + * or: 针对同一输入值，多个Predicate只要有一个返回True则返回True，否则返回False + */ + Assert.assertTrue(p.or(g).test("ta")); + } +} diff --git a/src/main/java/com/algorithm/study/demo/leetcode/Solution.java b/src/main/java/com/algorithm/study/demo/leetcode/Solution.java new file mode 100644 index 0000000..6a4253e --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/leetcode/Solution.java @@ -0,0 +1,28 @@ +package com.algorithm.study.demo.leetcode; + +/** + * 二分查找相关题目 + */ +public class Solution { + public static void main(String[] args) { + + } + /** + * 在有重复数字的有序数组中寻找n + * 假设按照升序排序的数组在预先未知的某个点上进行了旋转。 + * ( 例如，数组 [0,0,1,2,2,5,6] 可能变为 [2,5,6,0,0,1,2] )。 + * 编写一个函数来判断给定的目标值是否存在于数组中。若存在返回 true，否则返回 false。 + * 示例 1: + * 输入: nums = [2,5,6,0,0,1,2], target = 0输出: true + * 示例 2: + * 输入: nums = [2,5,6,0,0,1,2], target = 3输出: false + * @param nums + * @param target + * @return + */ + public boolean search(int[] nums, int target) { + int length=nums.length; + + return false; + } +} diff --git a/src/main/java/com/algorithm/study/demo/mode/decorator/Beverage.java b/src/main/java/com/algorithm/study/demo/mode/decorator/Beverage.java new file mode 100644 index 0000000..ff8e386 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/mode/decorator/Beverage.java @@ -0,0 +1,14 @@ +package com.algorithm.study.demo.mode.decorator; + +/** + * @Author: liuxun + * @CreateDate: 2019/2/7 下午5:13 + * @Version: 1.0 + */ +public abstract class Beverage { + protected String description=""; + public String getDescription(){ + return description; + } + public abstract double cost(); +} diff --git a/src/main/java/com/algorithm/study/demo/mode/decorator/Client.java b/src/main/java/com/algorithm/study/demo/mode/decorator/Client.java new file mode 100644 index 0000000..7de20da --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/mode/decorator/Client.java @@ -0,0 +1,13 @@ +package com.algorithm.study.demo.mode.decorator; + +/** + * @Author: liuxun + * @CreateDate: 2019/2/7 下午5:24 + * @Version: 1.0 + */ +public class Client { + public static void main(String[] args) { + Beverage beverage=new Milk(new Mocha(new HouseBlend())); + System.out.println(beverage.getDescription()+",cost:"+beverage.cost()); + } +} diff --git a/src/main/java/com/algorithm/study/demo/mode/decorator/CondimentDecorator.java b/src/main/java/com/algorithm/study/demo/mode/decorator/CondimentDecorator.java new file mode 100644 index 0000000..e73ccf2 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/mode/decorator/CondimentDecorator.java @@ -0,0 +1,11 @@ +package com.algorithm.study.demo.mode.decorator; + +/** + * 装饰类 + * @Author: liuxun + * @CreateDate: 2019/2/7 下午5:15 + * @Version: 1.0 + */ +public abstract class CondimentDecorator extends Beverage { + public abstract String getDescription(); +} diff --git a/src/main/java/com/algorithm/study/demo/mode/decorator/HouseBlend.java b/src/main/java/com/algorithm/study/demo/mode/decorator/HouseBlend.java new file mode 100644 index 0000000..46ff8f8 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/mode/decorator/HouseBlend.java @@ -0,0 +1,17 @@ +package com.algorithm.study.demo.mode.decorator; + +/** + * 具体的实现类 + * @Author: liuxun + * @CreateDate: 2019/2/7 下午5:13 + * @Version: 1.0 + */ +public class HouseBlend extends Beverage { + public HouseBlend(){ + description="House Bleand coffee"; + } + @Override + public double cost() { + return 4; + } +} diff --git a/src/main/java/com/algorithm/study/demo/mode/decorator/Milk.java b/src/main/java/com/algorithm/study/demo/mode/decorator/Milk.java new file mode 100644 index 0000000..1beda28 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/mode/decorator/Milk.java @@ -0,0 +1,24 @@ +package com.algorithm.study.demo.mode.decorator; + +/** + * 具体装饰类- + * @Author: liuxun + * @CreateDate: 2019/2/7 下午5:15 + * @Version: 1.0 + */ +public class Milk extends CondimentDecorator{ + protected Beverage beverage; + public Milk(Beverage beverage){ + this.beverage=beverage; + } + + @Override + public String getDescription() { + return beverage.getDescription()+",with milk"; + } + + @Override + public double cost() { + return 2+beverage.cost(); + } +} diff --git a/src/main/java/com/algorithm/study/demo/mode/decorator/Mocha.java b/src/main/java/com/algorithm/study/demo/mode/decorator/Mocha.java new file mode 100644 index 0000000..93be4a7 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/mode/decorator/Mocha.java @@ -0,0 +1,22 @@ +package com.algorithm.study.demo.mode.decorator; + +/** + * @Author: liuxun + * @CreateDate: 2019/2/7 下午5:17 + * @Version: 1.0 + */ +public class Mocha extends CondimentDecorator { + private Beverage beverage; + public Mocha(Beverage beverage){ + this.beverage=beverage; + } + @Override + public double cost() { + return 1+beverage.cost(); + } + + @Override + public String getDescription() { + return beverage.getDescription()+",with Mocha"; + } +} diff --git a/src/main/java/com/algorithm/study/demo/model/User.java b/src/main/java/com/algorithm/study/demo/model/User.java index ae11880..99395ee 100644 --- a/src/main/java/com/algorithm/study/demo/model/User.java +++ b/src/main/java/com/algorithm/study/demo/model/User.java @@ -6,7 +6,8 @@ public class User { private int id; private String name; - + public User(){ + } public User(int id,String name){ this.id=id; this.name=name; diff --git a/src/main/java/com/algorithm/study/demo/testng/Test.java b/src/main/java/com/algorithm/study/demo/testng/Test.java new file mode 100644 index 0000000..2f1f84a --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/testng/Test.java @@ -0,0 +1,11 @@ +package com.algorithm.study.demo.testng; + +import com.algorithm.study.demo.thread.SleepUtils; + +public class Test { + + @org.testng.annotations.Test(threadPoolSize = 10, invocationCount = 10000) + public void testJsf(){ + System.out.println("asdklfjalskfdj"+Thread.currentThread().getName()); + } +} diff --git a/src/main/java/com/algorithm/study/demo/thread/ThreadTest.java b/src/main/java/com/algorithm/study/demo/thread/ThreadTest.java new file mode 100644 index 0000000..29aeda9 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/thread/ThreadTest.java @@ -0,0 +1,7 @@ +package com.algorithm.study.demo.thread; + +public class ThreadTest { + public static void main(String[] args) { + + } +} diff --git a/src/main/java/com/algorithm/study/demo/util/DateUtils.java b/src/main/java/com/algorithm/study/demo/util/DateUtils.java new file mode 100644 index 0000000..dbca451 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/util/DateUtils.java @@ -0,0 +1,43 @@ +package com.algorithm.study.demo.util; + +import java.text.DateFormat; +import java.text.ParseException; +import java.text.SimpleDateFormat; +import java.util.Date; + +/** + * @title: DateUtils + * @projectName algorithm-study + * @description: TODO + * @date 2019/8/30 11:14 + */ +public class DateUtils { + private static final String formatStr = "HH:mm"; + public static void main(String args[]) throws ParseException { + String tS = "11:00"; + String tE = "12:10"; + System.out.println(getLong(tS)); + System.out.println(getLong(tE)); + System.out.println(getCurrentTime()); + System.out.println(isInZone(getLong(tS),getLong(tE),getCurrentTime())); + if(isInZone(getLong(tS),getLong(tE),getCurrentTime())){ + System.out.println("当前时间在范围中"); + }else{ + System.out.println("当前时间不在范围中"); + } + } + + private static boolean isInZone(long tStart,long tEnd,long t) throws ParseException { + return t>=tStart && t<=tEnd; + } + + private static long getLong(String timeStr) throws ParseException { + DateFormat dateFormat = new SimpleDateFormat(formatStr); + return dateFormat.parse(timeStr).getTime(); + } + + private static long getCurrentTime() throws ParseException { + DateFormat dateFormat = new SimpleDateFormat(formatStr); + return getLong(dateFormat.format(new Date())); + } +} diff --git a/src/main/java/com/algorithm/study/demo/util/GZIPUtils.java b/src/main/java/com/algorithm/study/demo/util/GZIPUtils.java new file mode 100644 index 0000000..15c1fe5 --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/util/GZIPUtils.java @@ -0,0 +1,225 @@ +package com.algorithm.study.demo.util; + + +import com.alibaba.fastjson.JSON; +import com.google.common.collect.Maps; +import org.apache.commons.io.FileUtils; + +import java.io.ByteArrayInputStream; +import java.io.ByteArrayOutputStream; +import java.io.File; +import java.io.IOException; +import java.util.Map; +import java.util.zip.*; + +/** + * @title: GZIPUtils + * @projectName algorithm-study + * @description: TODO + * @date 2019/10/17 20:50 + */ +public class GZIPUtils { + + /** + * 使用gzip进行压缩 + */ + public static String gzip(String primStr) { + if (primStr == null || primStr.length() == 0) { + return primStr; + } + + ByteArrayOutputStream out = new ByteArrayOutputStream(); + + GZIPOutputStream gzip = null; + try { + gzip = new GZIPOutputStream(out); + gzip.write(primStr.getBytes()); + } catch (IOException e) { + e.printStackTrace(); + } finally { + if (gzip != null) { + try { + gzip.close(); + } catch (IOException e) { + e.printStackTrace(); + } + } + } + + + return new sun.misc.BASE64Encoder().encode(out.toByteArray()); + } + + /** + *

Description:使用gzip进行解压缩

+ * + * @param compressedStr + * @return + */ + public static String gunzip(String compressedStr) { + if (compressedStr == null) { + return null; + } + + ByteArrayOutputStream out = new ByteArrayOutputStream(); + ByteArrayInputStream in = null; + GZIPInputStream ginzip = null; + byte[] compressed = null; + String decompressed = null; + try { + compressed = new sun.misc.BASE64Decoder().decodeBuffer(compressedStr); + in = new ByteArrayInputStream(compressed); + ginzip = new GZIPInputStream(in); + + byte[] buffer = new byte[1024]; + int offset = -1; + while ((offset = ginzip.read(buffer)) != -1) { + out.write(buffer, 0, offset); + } + decompressed = out.toString(); + } catch (IOException e) { + e.printStackTrace(); + } finally { + if (ginzip != null) { + try { + ginzip.close(); + } catch (IOException e) { + } + } + if (in != null) { + try { + in.close(); + } catch (IOException e) { + } + } + if (out != null) { + try { + out.close(); + } catch (IOException e) { + } + } + } + + return decompressed; + } + + /** + * 使用zip进行压缩 + * + * @param str 压缩前的文本 + * @return 返回压缩后的文本 + */ + public static final String zip(String str) { + if (str == null) { + return null; + } + byte[] compressed; + ByteArrayOutputStream out = null; + ZipOutputStream zout = null; + String compressedStr = null; + try { + out = new ByteArrayOutputStream(); + zout = new ZipOutputStream(out); + zout.putNextEntry(new ZipEntry("0")); + zout.write(str.getBytes()); + zout.closeEntry(); + compressed = out.toByteArray(); + compressedStr = new sun.misc.BASE64Encoder().encodeBuffer(compressed); + } catch (IOException e) { + compressed = null; + } finally { + if (zout != null) { + try { + zout.close(); + } catch (IOException e) { + } + } + if (out != null) { + try { + out.close(); + } catch (IOException e) { + } + } + } + return compressedStr; + } + + /** + * 使用zip进行解压缩 + * + * @param compressedStr 压缩后的文本 + * @return 解压后的字符串 + */ + public static final String unzip(String compressedStr) { + if (compressedStr == null) { + return null; + } + ByteArrayOutputStream out = null; + ByteArrayInputStream in = null; + ZipInputStream zin = null; + String decompressed = null; + try { + byte[] compressed = new sun.misc.BASE64Decoder().decodeBuffer(compressedStr); + out = new ByteArrayOutputStream(); + in = new ByteArrayInputStream(compressed); + zin = new ZipInputStream(in); + zin.getNextEntry(); + byte[] buffer = new byte[1024]; + int offset = -1; + while ((offset = zin.read(buffer)) != -1) { + out.write(buffer, 0, offset); + } + decompressed = out.toString(); + } catch (IOException e) { + decompressed = null; + } finally { + if (zin != null) { + try { + zin.close(); + } catch (IOException e) { + } + } + if (in != null) { + try { + in.close(); + } catch (IOException e) { + } + } + if (out != null) { + try { + out.close(); + } catch (IOException e) { + } + } + } + return decompressed; + } + + public static void main(String[] args) throws IOException { + //78910.txt 123456.txt + String strOld = FileUtils.readFileToString(new File("D:/78910.txt"), "utf-8"); + System.out.println("压缩前长度："+strOld.length()); + String gzip = zip(strOld); + System.out.println("压缩后长度："+gzip.length()); + String unzip = unzip(gzip); + FileUtils.write(new File("D:/2234567.txt"),unzip,"UTF-8"); +// int num=10000; +// +// long beginTime = System.currentTimeMillis(); +// for (int i = 0; i < num; i++) { +// zip(strOld); +// } +// long endTime = System.currentTimeMillis(); +// System.out.println("压缩总耗时"+(endTime - beginTime)); +// System.out.println("压缩平均耗时"+(endTime - beginTime) / 10000); +// +// long currentTimeMillis = System.currentTimeMillis(); +// for (int i = 0; i < 10000; i++) { +// unzip(gzip); +// } +// long endTimeMillis = System.currentTimeMillis(); +// System.out.println("解压总耗时"+(endTimeMillis - currentTimeMillis)); +// System.out.println("解压平均耗时"+(endTimeMillis - currentTimeMillis) / 10000); + + } +} diff --git a/src/main/java/com/algorithm/study/demo/util/JodaTimeUtil.java b/src/main/java/com/algorithm/study/demo/util/JodaTimeUtil.java new file mode 100644 index 0000000..04432fc --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/util/JodaTimeUtil.java @@ -0,0 +1,106 @@ +package com.algorithm.study.demo.util; + +import org.apache.commons.lang3.StringUtils; +import org.joda.time.DateTime; +import org.joda.time.format.DateTimeFormat; +import org.joda.time.format.DateTimeFormatter; + +import java.util.Date; + +/** + * @description: JodaTime工具类 + * @date 2019/5/31 11:24 + **/ +public class JodaTimeUtil { + public static final String STANDARD_FORMAT = "yyyy-MM-dd HH:mm:ss"; + public static final String STANDARD_DAY_FORMAT = "yyyy-MM-dd"; + public static final String STANDARD_DAY_FORMAT_1 = "yyyyMMdd"; + public static final String STANDARD_MILLIS_FORMAT="yyyy-MM-dd HH:mm:ss.SSS"; + public static final String STANDARD_MINUTE_FORMAT="yyyyMMddHHmm"; + + public static Date strToDate(String dateTimeStr,String formatStr){ + DateTimeFormatter dateTimeFormatter = DateTimeFormat.forPattern(formatStr); + DateTime dateTime = dateTimeFormatter.parseDateTime(dateTimeStr); + return dateTime.toDate(); + } + public static DateTime dateToDateTime(Date date){ + DateTime dateTime = new DateTime(date); + return dateTime; + } + public static Date strToDate(String dateTimeStr){ + DateTimeFormatter dateTimeFormatter = DateTimeFormat.forPattern(STANDARD_FORMAT); + DateTime dateTime = dateTimeFormatter.parseDateTime(dateTimeStr); + return dateTime.toDate(); + } + public static Date getNow(){ + return DateTime.now().toDate(); + } + public static String getNowFormat(String formatStr){ + return DateTime.now().toString(formatStr); + } + public static String getFormatNow(){ + return DateTime.now().toString(STANDARD_DAY_FORMAT_1); + } + public static String getFormatNowDay(){ + return DateTime.now().toString(STANDARD_DAY_FORMAT); + } + + public static Long strToMillis(String dateTimeStr,String formatStr){ + DateTimeFormatter dateTimeFormatter = DateTimeFormat.forPattern(formatStr); + long dateTime = dateTimeFormatter.parseDateTime(dateTimeStr).getMillis(); + return dateTime; + } + + public static String dateToStr(Date date,String formatStr){ + if(date == null){ + return StringUtils.EMPTY; + } + DateTime dateTime = new DateTime(date); + return dateTime.toString(formatStr); + } + public static String dateStrToMinuteStr(String dateTimeStr,String formatStr1,String formatStr2){ + DateTimeFormatter dateTimeFormatter = DateTimeFormat.forPattern(formatStr1); + DateTime dateTime = dateTimeFormatter.parseDateTime(dateTimeStr); + return dateTime.toString(formatStr2); + } + + + + public static String dateToStr(Date date){ + if(date == null){ + return StringUtils.EMPTY; + } + DateTime dateTime = new DateTime(date); + return dateTime.toString(STANDARD_FORMAT); + } + + /** + * 解析日期 yyyy-MM-dd HH:mm:ss + * + * @param timestamp + * @return + */ + public static String format(Long timestamp, String pattern) { + String dateStr = ""; + if (null == timestamp || timestamp.longValue() < 0) { + return dateStr; + } + try { + Date date = new Date(timestamp); + dateStr=dateToStr(date,pattern); + } catch (Exception e) { + // ignore + } + return dateStr; + } + public static DateTime strToDateTime(String dateTimeStr,String formatStr){ + Date date = strToDate(dateTimeStr, formatStr); + return dateToDateTime(date); + } + + public static void main(String[] args) { + Date createTime = JodaTimeUtil.strToDate("2019-10-24 02:04:41.921", JodaTimeUtil.STANDARD_MILLIS_FORMAT); + DateTime dateTime = dateToDateTime(createTime); + System.out.println(dateTime.getHourOfDay()); + } +} diff --git a/src/main/java/com/algorithm/study/demo/util/Paging.java b/src/main/java/com/algorithm/study/demo/util/Paging.java index c1c6427..0c96ffe 100644 --- a/src/main/java/com/algorithm/study/demo/util/Paging.java +++ b/src/main/java/com/algorithm/study/demo/util/Paging.java @@ -72,4 +72,7 @@ public int getStart() { return Math.max((page - 1) * pageSize, 0); } + public static void main(String[] args) { + + } } diff --git a/src/main/java/com/algorithm/study/demo/util/ZipUtil.java b/src/main/java/com/algorithm/study/demo/util/ZipUtil.java new file mode 100644 index 0000000..765e78b --- /dev/null +++ b/src/main/java/com/algorithm/study/demo/util/ZipUtil.java @@ -0,0 +1,54 @@ +package com.algorithm.study.demo.util; + +/** + * @title: ZipUtil + * @projectName algorithm-study + * @description: TODO + * @date 2019/10/18 13:39 + */ +import java.io.ByteArrayInputStream; +import java.io.ByteArrayOutputStream; +import java.io.IOException; +import java.util.zip.GZIPInputStream; +import java.util.zip.GZIPOutputStream; + +// 将一个字符串按照zip方式压缩和解压缩 +public class ZipUtil { + + // 压缩 + public static String compress(String str) throws IOException { + if (str == null || str.length() == 0) { + return str; + } + ByteArrayOutputStream out = new ByteArrayOutputStream(); + GZIPOutputStream gzip = new GZIPOutputStream(out); + gzip.write(str.getBytes()); + gzip.close(); + return out.toString("ISO-8859-1"); + } + + // 解压缩 + public static String uncompress(String str) throws IOException { + if (str == null || str.length() == 0) { + return str; + } + ByteArrayOutputStream out = new ByteArrayOutputStream(); + ByteArrayInputStream in = new ByteArrayInputStream(str + .getBytes("ISO-8859-1")); + GZIPInputStream gunzip = new GZIPInputStream(in); + byte[] buffer = new byte[256]; + int n; + while ((n = gunzip.read(buffer)) >= 0) { + out.write(buffer, 0, n); + } + // toString()使用平台默认编码，也可以显式的指定如toString("GBK") + return out.toString(); + } + + // 测试方法 + public static void main(String[] args) throws IOException { + System.out.println(ZipUtil.compress("中国China")); + System.out.println(ZipUtil.uncompress(ZipUtil.compress("中国China"))); + } + +}