LeetCode刷题：461.Hamming Distance

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.

解题思路：

n = x ^ y，两数对应位置若不相同，则n的相应位置置1
利用n = n & (n - 1)获取n中1的个数

解答：

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