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.

解题：
寻找连续的N个数中缺失的那个数，那么直接用原来N个数的和减去新的array中的和，即可得出缺失的那个数。可以AC的C++代码如下:

    int missingNumber(vector<int>& nums) {        int size = nums.size();        if(size < 1)            return 0;        int sum = (size + 1) * size / 2;        int sum2 = 0;        for(int i =0; i<size; i++)            sum2 += nums[i];        return sum - sum2;    }