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HearenHearen
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Merge pull request Hearen#29 from zhewen166/master
week two
2 parents 4ff9e3e + c0e099d commit 5558791

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‎LiuCaiZheng/src/com/cz/algorithm/learn/two/FindRepeatDNASequence.java‎

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package com.cz.algorithm.learn.two;
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import java.util.ArrayList;
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import java.util.HashSet;
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import java.util.List;
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import java.util.Set;
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public class FindRepeatDNASequence {
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public static void main(String[] args) {
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String s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT";
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new FindRepeatDNASequence().findRepeatedDnaSequences(s);
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}
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/**
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* during the process to compute the subString, we can use the hash algorithm to compute the hash value to identify weather two string is equal or not
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* this is a inspect and method to solve the class question.
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* */
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public List<String> findRepeatedDnaSequences(String s) {
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List<String> result = new ArrayList<String>();
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if (s == null || s.length() < 10)
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return result;
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int len = s.length();
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Set<Integer> set = new HashSet<Integer>();
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Set<Integer> unique = new HashSet<Integer>();
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int hash = 0;
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for (int i = 0; i < len; i++) {
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if (i < 9) {
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hash = (hash << 2) + getVal(s.charAt(i));
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} else {
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hash = (hash << 2) + getVal(s.charAt(i));
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// in the subStirng it has 10 , so the 20bit is enough for the
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// subString.
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hash &= (1 << 20) - 1;
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if (set.contains(hash) && !unique.contains(hash)) {
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result.add(s.substring(i - 9, i + 1));
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unique.add(hash);
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System.out.println(s.substring(i - 9, i + 1));
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} else {
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set.add(hash);
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}
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}
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}
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return result;
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}
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private int getVal(char ch) {
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if (ch == 'A')
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return 0;
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if (ch == 'C')
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return 1;
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if (ch == 'G')
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return 2;
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if (ch == 'T')
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return 3;
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return -1;
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}
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}

‎LiuCaiZheng/src/com/cz/algorithm/learn/two/LargestRectangleinHistogram.java‎

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package com.cz.algorithm.learn.two;
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import java.util.Stack;
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public class LargestRectangleinHistogram {
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// public static void main(String[] args) {
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// int[] height = new int[] {2,1,5,6,2,3};
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// new LargestRectangleinHistogram().largestRectangleArea(height);
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// }
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public int largestRectanglaArea2(int[] height) {
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return 0;
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}
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/**
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* the methodology is to create a stack which store the height of the array,
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* if the height[i] is greater or equal to the top of the stack , then push
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* height[i] into the stack else compute the max area , on the other hand,
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* record the number of which pop from the stack and push itself to the
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* stack when the process ends namely reaching the end of the array , pop
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* the stack until empty ,computing the area. at last, return the max area.
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*
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* the time complexity is O(n2); the space complexity is O(n);
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*
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* */
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public int largestRectangleArea(int[] height) {
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if (height == null || height.length == 0)
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return 0;
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int len = height.length;
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if (len == 1)
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return height[0];
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int max = 0;
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Stack<Integer> stack = new Stack<Integer>();
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int i = 0;
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while (i < len) {
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if (stack.empty() || height[i] >= stack.peek()) {
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stack.push(height[i]);
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i++;
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} else {
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int count = 0;
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while (!stack.empty() && stack.peek() > height[i]) {
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count++;
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int top = stack.peek();
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stack.pop();
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max = Math.max(max, top * count);
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}
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for (int j = 0; j < count + 1; j++) {
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stack.push(height[i]);
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}
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i++;
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}
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}
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int count = 0;
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while (!stack.empty()) {
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count++;
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max = Math.max(max, stack.peek() * count);
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stack.pop();
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}
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return max;
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}
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}

‎LiuCaiZheng/src/com/cz/algorithm/learn/two/LongestSubstring.java‎

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package com.cz.algorithm.learn.two;
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import java.util.HashMap;
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import java.util.Map;
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public class LongestSubstring {
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public static void main(String[] args) {
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String s = "asdfgfsds";
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int n = findLongestSubString(s);
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System.out.println(n);
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}
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/**
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* 暴力破解这个解法,时间复杂度是 O(n2).
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* 不是最优解.不可取.
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*
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* 把字符串转化为字符数组,对字符数组排序,
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* 然后一遍扫描, 时间复杂度取决于排序算法 最好的是 O(nLogn)
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*
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* 应该存在时间复杂度是O(n)的情况
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* 使用map数据结构进行解决.
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*
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* 思路说明:寻找两个相同字母之间的长度,作为嫌疑长度进行考察.
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* 并且保存一个中间值进行标记.
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* */
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public static int findLongestSubString(String s) {
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if(s == null || s.length() == 0) return 0;
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int max = -1;
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int lastIndex = -1;
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Map <Character, Integer> letterMap = new HashMap<Character, Integer>();
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for(int i = 0 ; i < s.length(); i ++) {
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char temp = s.charAt(i);
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if(letterMap.containsKey(temp) && lastIndex < letterMap.get(temp)) {
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lastIndex = letterMap.get(temp);
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}
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if( i - lastIndex > max)
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max = i - lastIndex;
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letterMap.put(temp , i);
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}
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return max;
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}
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}

‎LiuCaiZheng/src/com/cz/algorithm/learn/two/MaxPoints.java‎

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package com.cz.algorithm.learn.two;
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import java.util.HashMap;
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import java.util.Map;
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public class MaxPoints {
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public int maxPoints(Point[] points) {
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if(points == null) return 0;
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int len = points.length;
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if (len < 3)
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return len;
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int result = 0;
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Map<Double, Integer> slopeMap = new HashMap<Double, Integer>();
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for (int i = 0; i < len; i++) {
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slopeMap.clear();
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int duplicate = 1;
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double slope = 0.0;
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for (int j = 0; j < len; j++) {
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if (i == j) continue;
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if (points[i].x == points[j].x && points[i].y == points[j].y) {
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duplicate++;
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continue;
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} else if (points[i].x == points[j].x) {
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slope = Integer.MAX_VALUE;
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} else {
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slope = 1.0 * (points[i].y - points[j].y) / (points[i].x - points[j].x);
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}
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slopeMap.put(slope, slopeMap.containsKey(slope) ? slopeMap.get(slope) + 1 : 1);
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}
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if(slopeMap.keySet().size() == 0) {
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result = duplicate > result ? duplicate : result;
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} else {
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for (double key : slopeMap.keySet()) {
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result = Math.max((duplicate + slopeMap.get(key)), result);
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}
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}
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}
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return result;
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}
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class Point {
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int x;
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int y;
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public Point() {
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x = 0;
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y = 0;
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}
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public Point(int x, int y) {
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super();
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this.x = x;
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this.y = y;
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}
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}
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}

‎LiuCaiZheng/src/com/cz/algorithm/learn/two/MaximalRectangle.java‎

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package com.cz.algorithm.learn.two;
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import java.util.Stack;
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public class MaximalRectangle {
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public static void main(String[] args) {
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char[][] matrix = { { '0', '1', '1', '0', '1' },
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{ '1', '1', '0', '1', '0' }, { '0', '1', '1', '1', '0' },
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{ '1', '1', '1', '1', '0' }, { '1', '1', '1', '1', '1' },
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{ '0', '0', '0', '0', '0' } };
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System.out.println(new MaximalRectangle().maximalRectangle(matrix));
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}
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/**
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* here we apply the methodology of the last topic and transform the
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* question to this circumstance but how we transform this topic. what i do
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* here is that the max continues sequence of the number 1 construct the
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* height of the histogram, next step is to use the method previous. so we
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* can solve the question through two step: (1) find the max continues
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* sequence of the number 1 (2) use the method previous (3) the proof of the
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* transformation is in the resources of the project. the file name is
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* MaximalRectangle.docx
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*
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* the time complexity of the method is O(N3) the space complexity is O(n);
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*
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* */
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public int maximalRectangle(char[][] matrix) {
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if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
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return 0;
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int m = matrix.length;
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int n = matrix[0].length;
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int max = 0;
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int[] height = new int[n];
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (matrix[i][j] == '0')
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height[j] = 0;
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else
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height[j] += 1;
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}
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max = Math.max(largestRectangleArea(height), max);
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}
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return max;
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}
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private int largestRectangleArea(int[] height) {
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if (height == null || height.length == 0)
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return 0;
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int len = height.length;
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if (len == 1)
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return height[0];
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int max = 0;
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Stack<Integer> stack = new Stack<Integer>();
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int i = 0;
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while (i < len) {
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if (stack.empty() || height[i] >= stack.peek()) {
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stack.push(height[i]);
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i++;
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} else {
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int count = 0;
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while (!stack.empty() && stack.peek() > height[i]) {
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count++;
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int top = stack.peek();
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stack.pop();
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max = Math.max(max, top * count);
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}
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for (int j = 0; j < count + 1; j++) {
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stack.push(height[i]);
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}
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i++;
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}
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}
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int count = 0;
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while (!stack.empty()) {
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count++;
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max = Math.max(max, stack.peek() * count);
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stack.pop();
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}
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return max;
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}
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}

‎LiuCaiZheng/src/com/cz/algorithm/learn/two/SudokuSolver.java‎

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package com.cz.algorithm.learn.two;
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public class SudokuSolver {
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public static void main(String[] args) {
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}
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public void solveSudoku(char[][] board) {
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solution(board);
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}
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private boolean solution(char[][] board) {
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for (int i = 0; i < 9; i++) {
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for (int j = 0; j < 9; j++) {
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if (board[i][j] == '.') {
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for (char k = '1'; k <= '9'; k++) {
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board[i][j] = k;
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if (isValid(board, i, j) && solution(board)) {
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return true;
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}
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board[i][j] = '.';
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}
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return false;
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}
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}
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}
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return true;
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}
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// check weather satisfy the condition
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private boolean isValid(char[][] board, int x, int y) {
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// TODO Auto-generated method stub
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for(int i = 0; i < 9 ; i ++) {
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if(y != i && board[x][y] == board[x][i] ) {
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return false;
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}
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}
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for(int i = 0 ; i < 9 ; i ++) {
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if(x != i && board[x][y] == board[i][y] ) {
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return false;
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}
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}
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for(int i = (x / 3) * 3 ; i < (x / 3) * 3 + 3 ; i ++) {
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for(int j = ( y / 3) * 3 ; j < ( y / 3) * 3 + 3 ; j ++ ) {
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if (x != i && y != j && board[i][j] == board[x][y]) {
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return false;
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}
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}
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}
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return true;
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}
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}

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