448. Find All Numbers Disappeared in an Array

448. Find All Numbers Disappeared in an Array

Easy

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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]:
        """
         Use the array subscript to mark whether the corresponding number appears 
         Time complexity ：O(n)  Spatial complexity ：O(1)
        """
        #  The worst time complexity here is 2n
        # for i in range(len(nums)):
        #     num = nums[i]
        #     while num > 0 and nums[num - 1] != 0:
        #         indexNum = nums[num - 1]
        #         nums[num - 1] = 0
        #         num = indexNum
        # result = []
        # for i in range(len(nums)):
        #     if nums[i] != 0:
        #         result.append(i + 1)
        # return result

        #  Reference others' optimization time is n
        for num in nums:
            index = abs(num) - 1
            if nums[index] > 0:
                #  Clever ： In order not to lose the original data at this location , Only negative means that there has been a number in this position , The original data can still be obtained by using the absolute value each time 
                nums[index] = -nums[index]
        result = []
        for i in range(len(nums)):
            if nums[i] > 0:
                result.append(i + 1)
        return result