LeetCode 461. Hamming Distance 汉明距离

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：

1. 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);

2. 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();
    }
}