461. 汉明距离

https://leetcode-cn.com/problems/hamming-distance/

#define LL long long
class Solution {
public:
    int hammingDistance(int x, int y) {
        LL t = x^y;
        LL ans=0;
        while(t){
            ans+=1;
            t = t&(t-1); 
        }
        return ans;
    }
};

补一张真值表

另外学到了一个 x & (x-1)的妙用
x-1会使得x的最右边的1变为0
x & (x-1) 假如x只有一个1，那么结果就是0了

class Solution {
public:
    int hammingDistance(int x, int y) {
        x = x ^ y;  //异或，只留下两者不同的位数为1
        y = 0;
        while(x){
            y++;
            x = x & (x-1);
        }
        return y;
    }
};

python的实现一样的

class Solution:
    def hammingDistance(self, x: int, y: int) -> int:
        x=x^y
        ans=0
        while(x):
            ans+=1
            x=x&(x-1)
        return ans

但是更快的是直接调库

class Solution:
    def hammingDistance(self, x: int, y: int) -> int:
        z=x^y
        return bin(z).count("1")

bin的文档解释：

bin(x)
Convert an integer number to a binary string. The result is a valid Python expression. If x is not a Python int object, it has to define an index() method that returns an integer.

来源：CSDN

作者：hhmy77

链接：https://blog.csdn.net/hhmy77/article/details/104698256