【leetcode】【c++】268. Missing Number

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题目：

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?

即为给一个含有 n 个元素的向量，包含了 从 0 到 n 共 n+1 个值中的n个数，求少的那个数，则只需要用 0 到 n 的和(n是向量的size可以很简单的得到)减去实际向量的所有元素的和，即可得到Missing Number

代码为：

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int sum = 0;
        for(int i = 0; i <= nums.size(); i++)
        {
            sum += i;
        }
        vector<int>::iterator it;
        int t = 0;
        for(it = nums.begin(); it != nums.end(); it++)
            t += *it;
        return sum - t;
    }
};