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?

代码实现：

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int sum=0;
        int realsum=0;
        bool flag=false;
        for(int i=0;i<nums.size();++i){
            sum+=i;
            realsum+=nums[i];
        }
        sum+=nums.size();
        return sum-realsum;
    }
};