題目：

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.

題目大意是給出輸入十進(jìn)制數(shù)，要計(jì)算他們二進(jìn)制的漢明距離（長(zhǎng)度相同的兩個(gè)字符串，它們對(duì)應(yīng)位置不同字符的個(gè)數(shù)就是它們的漢明距離）。

解題思路：

1. 首先將兩個(gè)數(shù)進(jìn)行位異或運(yùn)算，下面以1和4舉例說(shuō)明：

      1   (0 0 0 1)
      4   (0 1 0 0)
   位異或  (0 1 0 1)

由上可知1和4進(jìn)行位異或運(yùn)算的結(jié)果是(0 1 0 1);

2. 然后計(jì)算上面結(jié)果中1的個(gè)數(shù)就是漢明距離了，完整代碼如下：

public class Solution {
    public int hammingDistance(int x, int y) {
        return Integer.toBinaryString(x^y).replace("0","").length();
    }
}