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…

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runtime?…

题目链接: https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/ 题目描述: Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] incl…

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 方法一:暴力求解 方法二:原地变负做标记 方法三:使用set 日期 [LeetCode] 题目地址:https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/ Total Accepted: 14302 Total Submissions: 24993 Difficulty:…

My first reaction is to have an unlimited length of bit-array, to mark existence. But if no extra mem is allowed, we can simply use 'sign' on each index slot to mark the same info.. it is a common technique. class Solution { public: vector<int> find…

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runtime?…

题目要求 Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) run…

problem 448. Find All Numbers Disappeared in an Array solution: class Solution { public: vector<int> findDisappearedNumbers(vector<int>& nums) { vector<int> res; ; i<nums.size(); i++) { ; nums[tmp] = nums[tmp]> ? -nums[tmp] : n…

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runtime?…

题目: Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runt…

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runtime?…

题目 给定一个范围在 1 ≤ a[i] ≤ n ( n = 数组大小 ) 的 整型数组,数组中的元素一些出现了两次,另一些只出现一次. 找到所有在 [1, n] 范围之间没有出现在数组中的数字. 您能在不使用额外空间且时间复杂度为O(n)的情况下完成这个任务吗? 你可以假定返回的数组不算在额外空间内. 示例: 输入: [4,3,2,7,8,2,3,1] 输出: [5,6] 题解 时间:O(n) 空间:O(1)(可以优化,不用 tem 的交换两个元素的值) class Solution { pub…

Description Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O…

题目描述 给定n个数字的数组,里面的值都是1-n,但是有的出现了两遍,因此有的没有出现,求没有出现值这个数组中的值有哪些. 要求不能用额外的空间(除了返回列表之外),时间复杂度n 思路 因为不能用额外空间并且时间是O(n),所以不能用排序或者hash 通过在对应位置的值去确定下一个位置,一直到遍历完.其实自己也没有很成熟的思想. 比如对于[4,3,2,7,8,2,3,1]: A[0],4=>找到a[3],7=>a[6],3=>a[2]====注意有可能出现循环,行不通. 本来以为行不通,…

LeetCode--Find All Numbers Disappeared in an Array Question Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this arr…

题目: 给定一个整数数组,判断是否存在重复元素. Given an array of integers, find if the array contains any duplicates. 如果任何值在数组中出现至少两次,函数返回 true.如果数组中每个元素都不相同,则返回 false. Your function should return true if any value appears at least twice in the array, and it should return…

原题链接在这里:https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/ 题目: Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that d…

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runtime?…

这是悦乐书的第232次更新,第245篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第99题(顺位题号是448).给定一个整数数组,其中1≤a[i]≤n(n =数组的大小),一些元素出现两次,其他元素出现一次.找到[1,n]包含的所有元素,这些元素不会出现在此数组中.你可以在没有额外空间和O(n)运行时的情况下完成吗? 您可以假设返回的列表不计入额外空间.例如: 输入:[4,3,2,7,8,2,3,1] 输出:[5,6] 本次解题使用的开发工具是eclipse,jdk…

描述 Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runti…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/3712 访问. 给定一个范围在  1 ≤ a[i] ≤ n ( n = 数组大小 ) 的 整型数组,数组中的元素一些出现了两次,另一些只出现一次. 找到所有在 [1, n] 范围之间没有出现在数组中的数字. 您能在不使用额外空间且时间复杂度为O(n)的情况下完成这个任务吗? 你可以假定返回的数组不算在额外空间内. 输入:[4,3,2,7,8,2,3,1] 输出:[…

问题来源:Find All Numbers Disappeared in an Array 很久没有刷题了,感觉大脑开始迟钝,所以决定重拾刷题的乐趣.一开始不要太难,选一些通过率高的题目做,然后就看到了这个题目.我有些吃惊,这个题我虽然知道两种解法,但本身还是有难度的,居然通过率这么高.然后就搜索相关网页,看到一个和它很接近的题目<Find All Duplicates in an Array>,然后就释然了.这两个题目有相同的题干,只是问题略微不同,解法有相似之处.估计是因为题号太接近了,会…

1.题目描述 2.问题分析 使的 A[i] = i+1 ,最后检查不满足这个条件的i+1 .即为缺失的值. 3.代码 vector<int> findDisappearedNumbers(vector<int>& nums) { ; i < nums.size() ; ++i ){ ] ){ swap( nums[i] , nums[nums[i] - ] ); i--; } } vector<int> result ; ; i < nums.siz…

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…

https://leetcode.com/problems/find-all-duplicates-in-an-array/description/ 参考:http://www.cnblogs.com/grandyang/p/4843654.html Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all th…

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)…

287. Find the Duplicate Number   hard http://www.cnblogs.com/grandyang/p/4843654.html 51. N-Queens http://blog.csdn.net/linhuanmars/article/details/20667175  http://www.cnblogs.com/grandyang/p/4377782.html 思路就是利用一个pos[row]=col来记录 行号:row,Queen在第col列.…

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…

leetcode--217. 存在重复元素 题目描述:给定一个整数数组,判断是否存在重复元素. 如果存在一值在数组中出现至少两次,函数返回 true .如果数组中每个元素都不相同,则返回 false . 示例 1: 输入: [1,2,3,1] 输出: true 示例 2: 输入: [1,2,3,4] 输出: false /** * 思路:将参数中的数组循环遍历,一一添加到一个 HashSet 对象中,HashSet会过滤重复元素, * 只需要拿数组nums与添加了全部数组元素的set比较,大小不…