LeetCode461. Hamming Distance

文章目录

    • 一、题目
    • 二、题解

一、题目

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.
Example 2:

Input: x = 3, y = 1
Output: 1

Constraints:

0 <= x, y <= 231 - 1

二、题解

class Solution {
public:
    int hammingDistance(int x, int y) {
        int n = x ^ y;
        //计算n中1的个数
        n = (n & 0x55555555) + ((n >> 1) & 0x55555555);
        n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
        n = (n & 0x0f0f0f0f) + ((n >> 4) & 0x0f0f0f0f);
        n = (n & 0x00ff00ff) + ((n >> 8) & 0x00ff00ff);
        n = (n & 0x0000ffff) + ((n >> 16) & 0x0000ffff);
        return n;
    }
};

你可能感兴趣的:(数据结构,leetcode,c++,算法,位运算)