[Leetcode] 461. Hamming Distance 解题报告

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:

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.

思路：

Hamming距离其实就是二进制表示中，同位不同值的个数。这刚好就是异或操作结果的二进制表示中1的个数。所以我们先求异或值，然后数一下其中1的个数即可。

代码：

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