268. Missing Number

1.Question

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?

2.Code

class Solution {public:    int missingNumber(vector<int>& nums) {        int n = nums.size();        int count = 0;        int all = 0.5 * n * (n+1);        for(int i = 0; i < n; i++)        {            count += nums[i];        }        return all - count;    }};

3.Note

a.  遍历一次，将这些数加在一起，再把没有缺失的时候的和求出来，通过相减即得缺失值。