461. Hamming Distance Add to List

// 快速法求1的个数
int BitCount2(unsigned int n)
{
    unsigned int c =0 ;
    for (c =0; n; ++c)
    {
        n &= (n -1) ; // 清除最低位的1
    }
    return c ;
}

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.
//比较两个数二进制位的不同，输出不同的位数，  直接异或然后统计1的个数

 1 class Solution {
 2 public:
 3     int hammingDistance(int x, int y) {
 4         int z=x^y;
 5         int ans=0;
 6         while(z){
 7             if(z&1)
 8                 ans++;
 9             z>>=1;
10         }
11         return ans;
12     }
13 };

461. Hamming Distance Add to List

原文：http://www.cnblogs.com/hutonm/p/6236721.html