LeetCode--268. Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

Example 1:

Input: [3,0,1]
Output: 2

Example 2:

Input: [9,6,4,2,3,5,7,0,1]
Output: 8

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

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思路：

1.数学题，根据公式求出前n项和应该是多少再减去实际值即可。

附等差数列公式： 1) 第n项，an=a1+(n-1)d。

                                  2) 前n项和，Sn=n*a1+n(n-1)d/2或Sn=n(a1+an)/2

参考代码：

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