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…

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…

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…

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…

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 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…

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…

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…

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…

传送门 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 mu…

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…

aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAACRAAAAMMCAYAAAAhQhmZAAAMFGlDQ1BJQ0MgUHJvZmlsZQAASImVlwdUk8kWx+crKYQklEAEpITeBOlVauggIB1shCSEUGIIBBU7uqjg2kUUK7oqYlsLIGvFgoVFwF5fFFFZWRcLWFB5kwTQ574977zJmS+/3Ln3zn8mM9+ZAUDNhSMW56LqAOSJCiVxoYGslNQ0FkkGEPihAydgy…

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:寻找重复数[287] 题目描述 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) . 数组中只有一个重复的数字,但它可能不止重复出现一次.…

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…

287. 寻找重复数 这题的难点就在于下面的说明了,我们先不管下面的那些说明的要求,用常规的解法来解答下上的题目. 排序思想解法 先把原来的数组进行排序,然后逐个遍历,一旦发现后一个元素和当前的元素相等,那么就返回,这就是我们找到了重复数字.但是这种思想,就不满足说明里面的,不能改变原数组,虽然时间复杂度是满足O(n^2). 哈希思想 用个哈希集合(HashSet)来记录已经出现过的元素,一旦遍历到了元素曾经出现在集合当中,那么就返回,这就是需要寻找的重复数字. 重新排序的思想 这种思想说起来有…

今天分享的题目来源于 LeetCode 第 287 号问题:寻找重复数. 题目描述 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) . 数组中只有一个重复…

287. 寻找重复数 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) . 数组中只有一个重复的数字,但它可能不止重复出现一次. PS: 快慢指针思想, f…

寻找重复数 根据题意,数组中的数字都在1~n之间,所以数字的范围是小于数组的范围的,数组的元素可以和数组的索引相联系. 例如:nums[0] = 1 即可以将nums[0]作为索引 通过nums[0] 可以访问到nums[1],以此类推. 如左图所示,环的入口就是重复元素. 那么问题就转化为了如何找到入环的第一个节点的问题.时间复杂度为O(n) 慢指针可以定义为:nums[slow] 快指针可以定义为:nums[nums[fast]] class Solution { public int fi…

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…

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…

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,…

Question 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 …

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…

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…