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 = 4Output: 2Explanation:
1   (0 0 0 1)
4   (0 1 0 0)↑   ↑The above arrows point to positions where the corresponding bits are different.

题意：

给定两个整数x,y，计算x和y的汉明距离。汉明距离是指x、y的二进制表示中，相同位置上数字不相同的所有情况数。

public class Solution {public int hammingDistance(int x, int y) {int notSame = x ^ y;int count = 0;while (notSame != 0) {if (notSame % 2 == 1)++count;notSame /= 2;}return count;}
}

注意记得用 异或
0 ^ 1 = 1

1 ^ 0 = 1

1 ^ 1 = 0

0 ^ 0 = 0

然后就是我们平时10进制的话，每次去掉末尾的数字是 / 10,二进制的话就是／2，切记。