268. Missing Number 难度：medium

题目：

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.

思路：

等差数列求和，再逐个减去数组的每个元素。

程序：

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