7/10 每日一題(缺失值)

 Given an array nums of n integers where nums[i] is in the range [1, n], return an array of all the integers in the range [1, n] that do not appear in nums.

 

Example 1:

Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]

Example 2:

Input: nums = [1,1]
Output: [2]

 

Constraints:

• n == nums.length
• 1 <= n <= 105
• 1 <= nums[i] <= n

 

Follow up: Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

class Solution:

    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:

        res=[]

        for n in nums:

            in_dex=abs(n)-1 #當成索引

            if nums[in_dex] > 0 : #數字轉成負數判斷重複

                nums[in_dex]= - nums[in_dex]

       

        for i in range(len(nums)):

            if nums[i] > 0: #沒有轉成負數就表示缺失

                res.append(i+1) #索引是從0開始,但數組是從1開始

        return res