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)矩屁，要計(jì)算他們二進(jìn)制的漢明距離（長(zhǎng)度相同的兩個(gè)字符串豆瘫，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）杈女。
解題思路：
- 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);
- 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();
}
}
