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.

从0-N中取出不同的N-1个数，要我们返回这个缺失的数。

一共会有N-1对相同的数，所以可以使用亦或法来解。

class Solution {
    public int missingNumber(int[] nums) {
        if(nums.length==1)
            return nums[0]==0? 1:0;
        int result = nums.length;
        for(int i = 0 ;i<nums.length;i++)
        {
            result^=i^nums[i];
        }
        return result;
    }
}