LeetCode:461. Hamming Distance

 package BitManipulation;
 //Question 461. Hamming Distance
 /*
 The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
 Given two integers x and y, calculate the Hamming distance.
 Note:
 0 ≤ x, y < 231.
 Example:
 Input: x = 1, y = 4
 Output: 2
 Explanation:
 1   (0 0 0 1)
 4   (0 1 0 0)
        ↑   ↑
 The above arrows point to positions where the corresponding bits are different.
  */
 public class hammingDistance461 {
     public static int hammingDistance(int x, int y) {
         int count=0;
         while(x>0|y>0){
             count+=(x&1)^(y&1);
             x=x>>1;
             y=y>>1;
         }
         return count;
     }
     //test
     public static void main(String[] args){
         int x=1;
         int y=4;
         System.out.println(hammingDistance(x,y));
     }

     //study the solution of other people
     //example1:use Integer.bitCount
     public static int hammingDistance1(int x, int y) {
         return Integer.bitCount(x ^ y);
     }
     //example2:xor = (xor) & (xor-1)
     //4(100),1(001)->101&100=100->100&011=000
     //1000001001这样中间的0可以一步直接跳过，机智！
     public int hammingDistance2(int x, int y) {
         int count = 0;
         for(int xor = x^y; xor != 0; xor = (xor) & (xor-1)) count++;
         return count;
     }
 }