# 461. Hamming Distance

[ **461. Hamming Distance**](https://leetcode.com/problems/hamming-distance/description/)

The [Hamming distance](https://en.wikipedia.org/wiki/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; 
}
```
