LeetCode268之Missing Number

原题：

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?

class Solution {public:    int missingNumber(vector<int>& nums) {        int sum=0;        int n=nums.size();        for(auto &a:nums)        {            sum+=a;        }                return n*(n+1)/2-sum;    }};

技巧问题，也是智商问题。。。（此处为自己的智商着急一下）

等差数列求和一下，然后减掉数组的和就得到该missing number