[LeetCode]Missing Number

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?

首先对0～ n内数字做异或，在对数组内数字做异或。等价与在Single Number 1，在一堆重复数字中找1个。

class Solution {public:    int missingNumber(vector<int>& nums) {        int ret = 0;        for(int i=0; i<=nums.size(); ++i){            ret ^= i;        }        for(int i=0; i<nums.size(); ++i){            ret ^= nums[i];        }        return ret;    }};