[LeetCode] 461. Hamming Distance（位操作）

传送门

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:

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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.

思路

题意：给定两个数，求出其汉明距离

题解：将两个数异或，那么二进制位上相同的都变成0，不同变成1，统计一下有多少个1即可

 

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class Solution { public:     //6ms     int hammingDistance(int x, int y) {         int res = 0;         while (x && y){             if ((x & 1) != (y & 1)) res++;             x >>= 1;             y >>= 1;         }         while (x){             if (x & 1)  res++;             x >>= 1;         }         while (y){             if (y & 1)  res++;             y >>= 1;         }         return res;     }       //3ms     int hammingDistance(int x,int y){         x ^= y;         y = 0;         while (x){             y += (x & 1);             x >>= 1;         }         return y;     } };