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 solution:

public class Solution {
    public int hammingDistance(int x, int y) {
        int temp = x ^ y;
        int count = 0;
        for(int i = 0; i < 32; i++) {
            if ((temp & 1) == 1)
                count++;
            temp >>= 1;
        }
        return count;
    }
}

大牛的解法：

Paste_Image.png

此解法利用了异或的重要性质，即 A ^ A = 0.