268. Missing Number - Easy

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?

因为是由0~n组成的数组，只缺少一个数字，可以先通过公式计算出应有的和，再遍历数组一个个减去，剩下的数就是missing number

时间：O(N)，空间：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;
    }
}