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:
汉明距离指的是两个二进制编码中,有多少位数不同的数字
算法思路:
两列数据做异或,结果放在xor中,此时xor中1的个数就是汉明距离
xor的最低位与1做与运算,count++,然后数字右移,统计有多少个1
代码:
public class Solution {
public int hammingDistance(int x, int y) {
int count = 0;
int xor ;
xor = x ^ y;
while(xor > 0){
count += xor & 1;
xor = xor >> 1;
}
return count;
}
}