Missing Number - LetcCode 268

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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 Array Math Bit Manipulation

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 (H) First Missing Positive (M) Single Number (H) Find the Duplicate Number

分析：n个不同的整数，取值范围是0到n之间，要找出缺失的那个整数。
首先可以计算出0到n的和，然后数组的和也是可以求出来的，因此，两者相减就是缺失的那个数。

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