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

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