leetcode No461. Hamming Distance

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Question

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.

Algorithm

可以一位一位的去比，我这里用异或表示，然后统计二进制中1的个数

Accepted Code

class Solution {
public:
    int hammingDistance(int x, int y) {
        int t = x^y;
        int res = 0;
        while(t){
            if(t & 0x01 == 1)
                res++;
            t = t >> 1;
        }
        return res;
    }
};

换一种写法：

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