Contents.swift

/**
 Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

  

 Example 1:
 Input: nums = [3,0,1]
 Output: 2
 Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
 
 Example 2:
 Input: nums = [0,1]
 Output: 2
 Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.
 
 Example 3:
 Input: nums = [9,6,4,2,3,5,7,0,1]
 Output: 8
 Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.
  

 Constraints:
 - n == nums.length
 - 1 <= n <= 104
 - 0 <= nums[i] <= n
 - All the numbers of nums are unique.
  

 Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
 */
class Solution {
    func missingNumber(_ nums: [Int]) -> Int {
        var sum = nums.count * (nums.count + 1) / 2
        print(sum)
        for e in nums {
            sum -= e
        }
        return sum
    }
}

let s = Solution()
let r = s.missingNumber([9,6,4,2,3,5,7,0,1])
print(r)