题目描述：

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

解题思路：

题目与Single Number题目本质上是一样的。就是一堆数字里面，只有一个数字是单的，其他都是成双的，找出那个单着的数字。运用异或就行了。

代码如下：

public class Solution {

    public int missingNumber(int[] nums) {

        int length = nums.length;

        int res = 0;

        for(int i = 1; i <= length; i++)

            res ^= i;

        for(int i = 0; i < length; i++)

            res ^= nums[i];

        return res;

    }

}