https://leetcode.com/problems/hamming-distance/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:先将x与y异或,得到temp,其二进制表示中的"1"的个数即为海明距离.将temp向右移i位(i从0到30)然后和"1"与操作,即检测最右一位是否为1,是的话count++.

class Solution {
public:
    int hammingDistance(int x, int y) {
        int count = 0;
        int temp = x ^ y;
        for(int i = 0; i <= 30; i++){
            count += (temp >> i) & 1;
        }
        return count;
    }
};