461. Hamming Distance

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每日一题 2019 - 02 - 25

题目：

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:

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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.

解法：

这个题挺水的，让我们找出任意两个数字的二进制中不同的位置的数目，最简单的方法就是求两个数字的异或，得到的结果为1的位置的个数就是要求的答案。

代码：

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