461. Hamming Distance

461. Hamming Distance

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.

My Solutions:

用xor找到x和y不一样的位数，一直>> 1这个数字直到变成0

class Solution {
    public int hammingDistance(int x, int y) {
        int A = x ^ y;
        int cnt = 0;
        
        while(A != 0) {
            cnt += 1 & A;
            A = A >> 1;
        }
        return cnt;
    }
}

//因为int总共有32位，也可以这样写：
for(int i = 0; i < 32; i++){
    count += (xor >> i) & 1; 
}