欣迪

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?

var missingNumber = function(nums) {
    const Arr = new Array(nums.length + 1).fill(0)
    for(let i = 0; i < nums.length; i++) {
        Arr[nums[i]] = 1
    }
    return Arr.indexOf(0)
};

這題其實算很簡單,做一個陣列比輸入的陣列長度多一,再把已經有的數字一個一個填入。
最後缺漏的陣列索引就是答案。

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作者介紹 - 欣迪

欣迪

從設計到寫程式,發現自己有追求前端技巧的自虐傾向。不斷的踩坑,再從坑裡爬出來,慢慢對攀岩有點心得。 目前在多間公司擔任網站設計顧問。 同時也是網站架設公司負責人。

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