Missing Number

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?

目的就是判断缺失哪一个数字，，最简单的方法就是排好序，看看却哪一个。

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int n;
        for( n=0;n<nums.size();n++)
        {
            if(n!=nums[n])
            {
                return n;
            }
        }
        return n;
    }
};

最简单的方法：

等差数列前n项和 - 数组之和 即为所求

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