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

思路：
(1)先求x^y的结果res。
(2)再依次求32位res的每一位与1进行与操作的结果，若不为0，则Hamming Distance加一。
(3)每判断完一位，res右移一位继续判断下一位。

public class Solution {  
    public int hammingDistance(int x, int y) {  
        int res = x ^ y;  
        int count = 0;  
        for (int i = 0; i < 32; i++) {  
            if ((res & 1) != 0)  
                count++;  
            res >>= 1;  
        }  
        return