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.

这种题真的会有各种技巧，先上个自己写的：

class Solution {
public:
    int hammingDistance(int x, int y) 
    {
        int z = x ^ y;
        int res = 0;
        while (z != 0)
        {
            int mod = z % 2;
            res += mod;
            z >>= 1;
        }
        return res;
    }
};

贴一个优化过的：

class Solution {
public:
    int hammingDistance(int x, int y) {
        int z = x^y;
        int count = 0;
        while(z){
            count += z&1;
            z >>= 1;
        }
        return count;
    }
};