题目：
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 < 2 ^31.
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.
题目大意是给出输入十进制数，要计算他们二进制的汉明距离（长度相同的两个字符串，它们对应位置不同字符的个数就是它们的汉明距离）。
解题思路：
- 首先将两个数进行位异或运算，下面以1和4举例说明：
1 (0 0 0 1)
4 (0 1 0 0)
位异或 (0 1 0 1)
由上可知1和4进行位异或运算的结果是(0 1 0 1);
- 然后计算上面结果中1的个数就是汉明距离了，完整代码如下：
public class Solution {
public int hammingDistance(int x, int y) {
return Integer.toBinaryString(x^y).replace("0","").length();
}
}
