【leetcode】461. Hamming Distance

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, return the Hamming distance between them.

 

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.

 

1、利用mask 统计每一位上的数字是否为1

class Solution {
public:
    int hammingDistance(int x, int y) {
        // xoring 之后统计每一位的一的个数
        int res=0;
        // int 型的数据 4个字节 最多32位的数据
        int z=x^y;
        for(int i=31;i>=0;--i){
            int mask=(1<<i);
            if(z&mask){
                res++;
            }
        }
        return res;

    }
};

2、利用bulitn function （内建函数）直接统计每一位上1的个数

class Solution {
public:
    int hammingDistance(int x, int y) {
        
        int z=x^y;

        return __builtin_popcount(z);
 
    }
};