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.

原问题可以装换为，求二进制数中1的个数问题。

统计一个二进制数中 1 的个数。之前在编程之美中看到 用这种方法判断 ：下面while循环求二进制中1的个数，原理就是：  

一个数减去1，则这个数的二进制数中最后一个1及其后的数字取反。

x & (x - 1) 为它的二进制数中少一个1 的状态..

class Solution {
public:
    int hammingDistance(int x, int y) {
        int xy=x^y;
        int dis=0;
        while(xy)
        {
            xy=xy&(xy-1);
            dis++;
        }
        return dis;
    }
};