Description
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 1:
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.
分析:
1、要找汉明距离,也就是两个码字的对应位不同的数量
2、对x、y进行按位异或,结果就是对应位不同的位置为1,相同的为0
3、计算异或的结果含有1的个数
4、每次都是查看最低位是不是1,然后再无符号右移1位,再次查看下一位即可
5、int为32位查看32次
class Solution {
public int hammingDistance(int x, int y) {
int num = x ^ y;
int count = 0;
int moves = 32;
while((moves--) > 0){
if((num & 1) == 1){
count++;
}
num = num >>> 1;
}
return count;
}
}