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。