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

Statement

简体中文English

Metadata

• Link: 数组中重复的数据
• Difficulty: Medium
• Tag: 数组 哈希表

给你一个长度为 n 的整数数组 nums ，其中 nums 的所有整数都在范围 [1, n] 内，且每个整数出现 一次 或 两次 。请你找出所有出现 两次 的整数，并以数组形式返回。

你必须设计并实现一个时间复杂度为 O(n) 且仅使用常量额外空间的算法解决此问题。

 

示例 1：

输入：nums = [4,3,2,7,8,2,3,1]
输出：[2,3]

示例 2：

输入：nums = [1,1,2]
输出：[1]

示例 3：

输入：nums = [1]
输出：[]

 

提示：

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

Metadata

• Link: Find All Duplicates in an Array
• Difficulty: Medium
• Tag: Array Hash Table

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

Python 3

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

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最后更新: October 11, 2023