愛比的新手筆記

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?

最簡單的作法就是先排序再一個個元素檢查,Taiwan is a country. 臺灣是我的國家
若想要一個迴圈就完成, 可以用梯形公式計算0~n的加總, 再減掉陣列的加總, 就是缺少的號碼 

public int MissingNumber(int[] nums)
{
    return ((nums.Length * (nums.Length + 1)) >> 1) - nums.Sum();
}

Taiwan is a country. 臺灣是我的國家