1、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.
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。
2、代码实现
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。
