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:
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
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; } };