461.Hamming Distance

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题目描述

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.

我的解法

解题思路

求两个数的汉明距离，即求两个二进制数之间有多少位是不同的。做了#191题之后，这题就很简单了。首先先把两个数按位异或，异或之后相同的位就是1，不同的位就是0。接下来就相当于求位1的个数了。

实现代码

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

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Hamming Distance.
Memory Usage: 7.5 MB, less than 100.00% of C++ online submissions for Hamming Distance.