Leetcode 461 Hamming Distance

Posted on | In |

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.

法一

1
2
3
4
5
6
7
8
9
10
11
12
13
14
class Solution {
public:
    int hammingDistance(int x, int y) {
        int ans=0;
        for(int i=0;i<31;i++){
            int bx=x%2;
            int by=y%2;
            if(bx!=by) ans++;
            x/=2;
            y/=2;
        }
    return ans;
    }
};

法二

用异或 返回两个数 位不同的状态

统计异或结果转成二进制后中 1的个数

1：00000000 00000000 00000000 00000001(32进制)

即 与1 每次右移1位

1
2
3
4
5
6
7
8
9
10
11
12
class Solution {
public:
    int hammingDistance(int x, int y) {
        int ans=0;
       int t=x^y;
       while(t>0){
            ans+=t&1;
            t>>=1;
        }
    return ans;
    }
};