Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. Note: You must not modify th…

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate element must exist. Assume that there is only one duplicate number, find the duplicate one. Note: You must not modify t…

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. Note: You must not modify th…

Difficulty:medium  More:[目录]LeetCode Java实现 Description Given an array nums containing n + 1 integers where each integer is between 1 and n(inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate numbe…

LeetCode 287. Find the Duplicate Number 暴力解法 时间 O(nlog(n)),空间O(n),按题目中Note"只用O(1)的空间",照理是过不了的,但是可能判题并没有卡空间复杂度,所以也能AC. class Solution: # 基本思路为,将第一次出现的数字 def findDuplicate(self, nums: List[int]) -> int: s = set() for i in nums: a = i in s if a…

Leetcode之二分法专题-287. 寻找重复数(Find the Duplicate Number) 给定一个包含 n + 1 个整数的数组 nums,其数字都在 1 到 n 之间(包括 1 和 n),可知至少存在一个重复的整数.假设只有一个重复的整数,找出这个重复的数. 示例 1: 输入: [1,3,4,2,2] 输出: 2 示例 2: 输入: [3,1,3,4,2] 输出: 3 说明: 不能更改原数组(假设数组是只读的). 只能使用额外的 O(1) 的空间. 时间复杂度小于 O(n2)…

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. Example 1: Input: [1,3,4,2,2…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 保存已经访问过的数字 链表成环 二分查找 日期 题目地址:https://leetcode.com/problems/find-the-duplicate-number/description/ 题目描述 Given an array nums containing n + 1 integers where each integer is betwe…

最容易想到的思路是新开一个长度为n的全零list p[1~n].依次从nums里读出数据,假设读出的是4, 就将p[4]从零改成1.如果发现已经是1了,那么这个4就已经出现过了,所以他就是重复的那个数.这个解法的时间复杂度是O(N).但是由于本题要求空间复杂度是O(1).所以不能用. 可以用二分法,low = 1, high = n, mid = (left + right)//2,如果<=mid 的元素个数 > mid,那么重复的数字一定在[1, mid]区间内.反之,则一定在[mid+1,…

Description: Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. Note: You must…