问题描述:

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 ≤ xy < 231.

代码如下: 3ms 

int hammingDistance(int x, int y) 
{
    int z = x ^ y;
    
    int count = 0;
    
    while(z)
    {
        if(z % 2)
        {
            ++count;
        }
        
        z = z >> 1;
    }
    
    return count;
}

  

posted on 2017-12-05 11:03  gtxvs  阅读(132)  评论(0编辑  收藏  举报