LeetCode Array Easy 268. Missing Number

 Description

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?

 问题描述，一个数组包含n个不重复的数从0,1,2,3,...n 找到那个不在数组中的数

我的思路：先将数组排序再进行比较.但是这样子效率很低，时间复杂度在O=n+排序所需时间；

public int MissingNumber(int[] nums) {
        Array.Sort(nums);
       
        for(int i=0; i < nums.Length; i++){
            if(i != nums[i])
                return i;
        }
        return nums.Length;
    }

第二种思路： 从0到n的和为(0+n)(n+1)/2 再计算数组的和。两和的差就是结果。该算法时间复杂度为O(n) 

public int MissingNumber(int[] nums) {
        int n = nums.Length;
        int sum = n*(n+1)/2;
        int sums = nums.Sum();
        return sum-sums;
    }