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?

因為是由0~n組成的陣列，只缺少一個數字，可以先通過公式計算出應有的和，再遍歷陣列一個個減去，剩下的數就是missing number

時間：O(N)，空間：O(1)

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