LeetCode #268 - Missing Number

题目描述：

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?

给定[0,n]范围内的n个不同的数，其中只有一个数缺失了，需要在O(1)的时间复杂度和空间复杂度内找到它。由于0~n的总和已知，可以计算这n个数的和，那么差值就是缺失的那个数。

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