# 442.find-all-duplicates-in-an-array

## Statement

``````输入：nums = [4,3,2,7,8,2,3,1]

``````

``````输入：nums = [1,1,2]

``````

``````输入：nums = [1]

``````

• `n == nums.length`
• `1 <= n <= 105`
• `1 <= nums[i] <= n`
• `nums` 中的每个元素出现 一次两次

Given an integer array `nums` of length `n` where all the integers of `nums` are in the range `[1, n]` and each integer appears once or twice, return an array of all the integers that appears twice.

You must write an algorithm that runs in `O(n) `time and uses only constant extra space.

Example 1:

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

Example 2:

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

Example 3:

``````Input: nums = [1]
Output: []
``````

Constraints:

• `n == nums.length`
• `1 <= n <= 105`
• `1 <= nums[i] <= n`
• Each element in `nums` appears once or twice.

## Solution

``````from typing import List

class Solution:
def findDuplicates(self, nums: List[int]) -> List[int]:
n = len(nums)
nums.append(n + 1)
res = []
for i in range(n):
x = abs(nums[i])
if nums[x] < 0:
res.append(x)
else:
nums[x] *= -1

return res
``````