对于一个长度为n+1的数组,其中每一个值的取值范围是[1,n],可以证明的是必然存在一个重复数字(抽屉原理),假设仅存在一个重复数字,找到他. 举例:输入:[1,3,4,2,1],输出:1 自己做的时候,要么时间复杂度到o(n2),要么需要额外的存储空间利用hashset,下面来分析一下leetcode上别人的算法吧. 方法一:通过图论中环的有关知识解决 public int findDuplicate(int[] nums) { // Find the intersection point o…

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. Example 1: Input: [1,3,4,2,2…

Given a binary tree, return all duplicate subtrees. For each kind of duplicate subtrees, you only need to return the root node of any oneof them. Two trees are duplicate if they have the same structure with same node values. Example 1: 1 / \ 2 3 / /…

一个长度为 n + 1 的整形数组,其中的数字都在 1 到 n 之间,包括 1 和 n ,可知至少有一个重复的数字存在.假设只有一个数字重复,找出这个重复的数字.注意:    不能更改数组内容(假设数组是只读的).    只能使用恒定的额外空间,即要求空间复杂度是 O(1) .    时间复杂度小于 O(n2)    数组中只有一个数字重复,但它可能不止一次重复出现.详见:https://leetcode.com/problems/find-the-duplicate-number/descri…

题意: 有一个含有n+1个元素的数组,元素值是在1-n之间的整数,请找出其中出现超过1次的数.(保证仅有1个出现次数是超过1的数) 思路: 方法一:O(nlogn).根据鸽笼原理及题意,每次如果<=k的数超过了k个,那么答案必定在[1,k].可以用二分枚举答案来解决. bool left(vector<int>& nums,int tar)//是否在左边 { ; ; i<nums.size(); i++) if(nums[i]<=tar) cnt++; return…

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…

［抄题］: 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…

Level:   Medium 题目描述: 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. Exampl…

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…

LeetCode 652: 寻找重复的子树 Find Duplicate Subtrees 题目: 给定一棵二叉树,返回所有重复的子树.对于同一类的重复子树,你只需要返回其中任意一棵的根结点即可. 两棵树重复是指它们具有相同的结构以及相同的结点值. Given a binary tree, return all duplicate subtrees. For each kind of duplicate subtrees, you only need to return the root nod…

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…

后面3个题都是限制在1-n的,所有可以不先排序,可以利用巧方法做.最后两个题几乎一模一样. 217. Contains Duplicate class Solution { public: bool containsDuplicate(vector<int>& nums) { int length = nums.size(); ) return false; sort(nums.begin(),nums.end()); ;i < length;i++){ ]) return tr…

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,…

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…

题目 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…

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…

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…

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array. For example, Given nums = [0, 1, 3] return 2. Note: Your algorithm should run in linear runtime complexity. Could you implement it u…

Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k. 这道题跟之前两道Contains Duplicate 包含重复值和Conta…

Given an array of integers and an integer k, return true if and only if there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and j is at most k. (Old Version) Given an array of integers and an i…

Given an array of integers, every element appears twice except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 本来是一道非常简单的题,但是由于加上了时间复杂度必须是O(n),并且空间复杂度为O(1)…

10.6 You have 10 billion URLs. How do you detect the duplicate documents? In this case, assume that "duplicate" means that the URLs are identical. 这道题让我们在一百亿个URL链接中寻找相同项,看这数据量简直吓尿了,如果每个URL链接平均100个字符的话,每个字符是4个字节,那么总共需要占4TB的空间,我们无法在内存中导入这么大的数据量.假如…

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 number must exist. Assume that there is only one duplicate number, find the duplicate one. Note: You must not modify th…

问题描述: 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 mod…

转载:http://www.cnblogs.com/grandyang/p/4756677.html Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array. For example,Given nums = [0, 1, 3] return 2. Note:Your algorithm should run in li…

本文给大家介绍如何在JAVA中实现随机无重复数字的功能.如果您是初学者的话,有必要看一看这篇文章,因为这个功能一般会在面试中遇到.包括我本人在招聘人员的时候也喜欢拿这个问题去问别人,主要看一看考虑问题的模式和基础知识如何. 希望这篇文章能给初次接触的朋友一些帮助,因为我曾接触过一些朋友要么写不出来,要么使用很平铺的思维方式去实现它. 一般有点开发经验的朋友都能实现这样的功能,只不过是效率上的问题.我们一般在面对这样的问题时,总会平铺直序的联想到,先生成一个数组,然后在一个循环中向数组中添加随机数…

1.2.3 Name That Number 命名那个数字 Time Limit: 1 Sec  Memory Limit: 64 MBSubmit: 183  Solved: 33[Submit][Status][Forum] Description 在威斯康辛州牛大农场经营者之中,都习惯于请会计部门用连续数字给母牛打上烙印.但是,母牛用手机时并没感到这个系统的便利,它们更喜欢用它们喜欢的名字来呼叫它们的同伴,而不是用像这个的语句"C'mon, #4734, get along.".…