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]:
        """
        利用数组下标标记对应数字是否出现
        时间复杂度：O(n) 空间复杂度：O(1)
        """
        # 这里时间复杂度最坏为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

        # 参考别人优化时间为n
        for num in nums:
            index = abs(num) - 1
            if nums[index] > 0:
                # 取巧：为了不丢失该位置上的原数据，只取反代表该位置出现过数字，每次取值的时候用绝对值仍能获取原数据
                nums[index] = -nums[index]
        result = []
        for i in range(len(nums)):
            if nums[i] > 0:
                result.append(i + 1)
        return result