add lintcode

Author: codingWang

src/com/duwei/lintcode/MergeNum.java

@@ -0,0 +1,39 @@
+package com.duwei.lintcode;
+
+import java.util.PriorityQueue;
+
+/**
+ * LintCode:791
+ * 给出n个数，现在要将这n个数合并成一个数，每次只能选择两个数a,b合并，每次合并需要消耗a+b的能量，输出将这n个数合并成一个数后消耗的最小能量。
+ */
+public class MergeNum {
+
+    public int mergeNum(int[] nums) {
+        PriorityQueue<Integer> mHeap = new PriorityQueue<>();//小根堆
+        int value = 0;
+
+        for (int num : nums) {//全部add进去构造堆
+            mHeap.add(num);
+        }
+
+        while (mHeap.size() > 1) {
+            int num1 = mHeap.remove();
+            int num2 = mHeap.remove();
+            int afterAdd = num1 + num2;
+            value += afterAdd;
+            mHeap.add(afterAdd);
+        }
+
+        return value;
+    }
+
+    public static void main(String[] args) {
+        int[] a = {1, 2, 3, 4};
+
+        MergeNum mergeNum = new MergeNum();
+        System.out.print(mergeNum.mergeNum(a));
+
+    }
+
+
+}

src/com/duwei/lintcode/PlusAb.java

@@ -0,0 +1,28 @@
+package com.duwei.lintcode;
+
+/**
+ * 不使用+运算符计算和
+ */
+public class PlusAb {
+
+    public int add(int a, int b) {
+        if ((a & b) == 0) {//无进位情况
+            return a | b;
+        }
+
+        int d = a ^ b;
+        int c = (a & b) << 1;//进位
+
+        while ((d & c) != 0){
+            c = c<<1;
+        }
+
+        return c;
+    }
+
+
+    public static void main(String[] args) {
+        System.out.print(new PlusAb().add(1,1));
+    }
+
+}

src/com/duwei/lintcode/ReversNum.java

@@ -0,0 +1,20 @@
+package com.duwei.lintcode;
+
+
+/**
+ * 翻转一个三位数的整数
+ */
+public class ReversNum {
+    public static void main(String[] args) {
+
+    }
+
+
+    public int reversNum(int num){
+
+
+
+        return 0;
+    }
+
+}

src/com/duweri/interview/BitCompute.java

@@ -1,37 +1,50 @@
 package com.duweri.interview;
+
 /**
- * 位运算Demo
- * @author 杜伟
- * 
+ * 位运算Demo
+ *
+ * @author 杜伟
  */
 public class BitCompute {
 
-	public static void main(String[] args) {
-		//异或
-		int a = 8;//1000
-		int b = 9;//1001
-		System.out.println("a^b = "+(a^b));
-		
-		int x = 15;//01111=1+2+4+8
-		int y = 23;//10111=1+2+4+16
-		System.out.println("x^y = "+(x^y));
-		
-		//与运算
-		System.out.println("x&y = "+(x&y));
-		
-		//或运算
-		System.out.println("x|y = "+ (x|y));
-		
-		//左右移运算
-		System.out.println("8>>2 = "+(8>>2));
-		System.out.println("8<<2 = "+(8<<2));
-		
-		//~		原码01000放进计算机用补码：01000，
-		//取反计算：10111（这是补码），输出源码（取反加1，符号位不变）：11001
-		System.out.println("~ = "+(~8));
-		
-		System.out.println("8>>>2="+(8>>>2));
-		
-	}
+    public static void main(String[] args) {
+        //异或
+        int a = 8;//1000
+        int b = 9;//1001
+        ////////////0001
+        System.out.println("a^b = " + (a ^ b));
+
+        int x = 15;//01111=1+2+4+8
+        int y = 23;//10111=1+2+4+16
+        /////////////11000=16+8=24
+        System.out.println("x^y = " + (x ^ y));
+
+        //与运算
+        //int x = 15;//01111=1+2+4+8
+        //int y = 23;//10111=1+2+4+16
+        ///////////////00111=4+2+1=7
+        System.out.println("x&y = " + (x & y));
+
+        //或运算
+        //int x = 15;//01111=1+2+4+8
+        //int y = 23;//10111=1+2+4+16
+        ///////////////11111=16+8+4+2+1=31
+        System.out.println("x|y = " + (x | y));
+
+        //左右移运算
+        //0000 1000>>0000 0010
+        //0000 1000<<0010 0000
+        System.out.println("8>>2 = " + (8 >> 2));//2
+        System.out.println("8<<2 = " + (8 << 2));//32
+        System.out.println("8<<2 = " + (8 << 5));//256
+        System.out.println("1<<32 = " + (1 << 32));//1
+
+        //~		原码01000放进计算机用补码：01000，
+        //取反计算：10111（这是补码），输出源码（取反加1，符号位不变）：11001
+        System.out.println("~ = " + (~8));
+
+        System.out.println("8>>>2=" + (8 >>> 2));
+
+    }
 
 }