leetcode——Missing Number

原题目：

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

.

Note:

Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
解析：题目要求在线性复杂度以及常量空间内解决，首先这是一个缺失一个元素的以0为第一项的等差数列，所以可以根据等差数列前n项和公式来解决，补齐缺项，所以要按n+1项来计算，故完整n+1项和为n（n+1），减去数组中的n项和即可得到缺失项的值，算法简单易实现。

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