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461. Hamming Distance

时间:2017-07-27 13:14:14      阅读:237      评论:0      收藏:0      [点我收藏+]

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.

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.

统计一个二进制数中 1 的个数。之前在编程之美中看到 用这种方法判断 :x = x&( x - 1)  

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

 

461. Hamming Distance

原文:http://www.cnblogs.com/pk28/p/7244253.html

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