LeetCode(461) Hamming Distance

Posted 2022-12-12 逆風的薔薇

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, 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.

分析

题目含义是求两个整数对应的二进制串中不同比特位值的个数。

可以转化为 求 （x^y）对应二进制串中1的个数。

代码

class Solution 
    public int hammingDistance(int x, int y) 
        return countNumOf1Bits(x ^ y);
    

    public int countNumOf1Bits(int n) 
        int count = 0;
        while (n > 0) 
            if ((n & 1) == 1) 
                ++count;
            
            n >>= 1;
        
        return count;