【268】 Missing Number

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?

思路:

因为是从0到n中找丢失的数，只要算出理论上从0到n的和再减去给出的所有数即为所求。

代码：

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