1.题目描述 2.问题分析 使用快慢指针方法判断链表是否有环,然后寻找环开始的节点. 3.代码 ListNode *detectCycle(ListNode *head) { if( head == NULL || head->next == NULL ){ return NULL ; } ListNode* fast = head; ListNode* slow = head; while( fast != NULL && slow != NULL ){ if( fast->…

1.题目描述 2.问题分析 使用快慢指针方法,一个快指针,一个慢指针,如果到某个时候,快指针追上了慢指针,则说明有环存在. 3.代码 bool hasCycle(ListNode *head) { if( head == NULL || head->next == NULL ) return false; ListNode* f = head->next; ListNode* s = head; while( f != NULL && s != NULL ){ if( s ==…

LeetCode解题报告:Linked List Cycle && Linked List Cycle II 1题目 Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? Linked List Cycle II Given a linked list, return the node w…

Problem: Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? https://oj.leetcode.com/problems/linked-list-cycle/ Problem II: Given a linked list, return the node where the cycle begins. If the…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space?   Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Fol…

Difficulty:medium  More:[目录]LeetCode Java实现 Description Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up:Can you solve it without using extra space? Intuiti…

1. Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. 思路:想法是利用两指针,一个每次移动一步,另一个每次移动两步,如果存在环则这两个指针一定会相遇(这里可以在纸上画一下,因为后一个指针移动比前一个指针快,当后一个指针在环中来到前一个指针的…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 解法一: 使用unordered_map记录当前节点是否被访问过,如访问过说明有环,如到达尾部说明无环. /** * Definition for singly-linked list. * struct ListNode { * int va…

Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? 解法一: 使用unordered_map记录当前节点是否被访问过,如访问过返回该节点,如到达尾部说明无环. /** * Definition for sing…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? (二)解题 本题大意:给定一个链表,判断链表里面是否成环.不能用辅助空间. 解题思路:利用快…

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 双指针 保存已经走过的路径 日期 [LeetCode] 题目地址:https://leetcode.com/problems/linked-list-cycle/ Total Accepted: 102417 Total Submissions: 277130 Difficulty: Easy 题目描述 Given a linked list, de…

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 双指针 set 日期 题目地址:https://leetcode.com/problems/linked-list-cycle-ii/description/ 题目描述 Given a linked list, return the node where the cycle begins. If there is no cycle, return n…

Problem link: http://oj.leetcode.com/problems/linked-list-cycle/ We set two pointers: the faster pointer goes two steps each iteration, and the slower one goes one step. If the two pointers meet some time, it follows that there is a loop; otherwise,…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up:Can you solve it without using extra space? 这个题还是蛮考验数学推理的,不过在前一个题的基础上还是能推出结果的.这是英文一段解释,非常有帮助. First Step: A…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 解决这个问题需要很巧妙的思路,一般我们会想到观察链表中是否出现过重复出现的节点.但是这样的空间的复杂度是O(n),比如用set来存储出现过的节点,然后看下一个节点是否存在于set中. 一个很巧妙的解决办法是设置两个指针,一个快指针和一个慢指针.快指针每次移动两步,慢指针每次移动一…

题目: Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 提示: 首先,题目中要求'without using extra space',指的是空间复杂度必须控制在O(1)内. 因此可以创建两个变量,先同时指向head,然后每一轮循环中,令其中一个变量沿链表向前“走”两步,另一个走“一步”,这样的话每一个循环后他们两者的距离差会…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up:Can you solve it without using extra space? 和前面那题类似,只不过这里要找到的是环开始的节点,看下面这张图(图是盗的): 当fast以及slow指针在z处相遇的时候,可以…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 判断一个链表是否存在环,维护快慢指针就可以,如果有环那么快指针一定会追上慢指针,代码如下: class Solution { public: bool hasCycle(ListNode *head) { ListNode * slow, * fast; slow = fast…

Problem link: http://oj.leetcode.com/problems/linked-list-cycle-ii/ The solution has two step: Detecting the loop using faster/slower pointers. Finding the begining of the loop. After two pointers meet in step 1, keep one pointer and set anther to th…

这是悦乐书的第176次更新,第178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35题(顺位题号是141).给定一个链表,确定它是否有一个循环. 本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用Java语言编写和测试. 02 理解题意 什么样结构的链表才算是拥有一个循环呢? 链表中某一节点的引用指向了当前链表中已经存在的另一节点时,此链表存在循环. ListNode L = new ListNode(1); Li…

题目描述: 不用辅助空间判断,链表中是否有环 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ /*一个指针走的快 一个指针走得慢*/ class Solution { public: bool hasCycle(ListNode *head) { if(head ==…

一种方法是用set存储出现过的指针,重复出现时就是有环: class Solution { public: bool hasCycle(ListNode *head) { set<ListNode*> st; ListNode *p = head; while(p) { if (st.find(p) != st.end()) return true; st.insert(p); p = p->next; } return false; } }; 另一种方法就是快慢指针,快指针一次走两步,…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? 思路 这题是Linked List Cycle的进阶版 Given a linked list, determine if it has a cycle in it. bool hasCycle(Li…

1. 链表 数组是一种顺序表,index与value之间是一种顺序映射,以\(O(1)\)的复杂度访问数据元素.但是,若要在表的中间部分插入(或删除)某一个元素时,需要将后续的数据元素进行移动,复杂度大概为\(O(n)\).链表(Linked List)是一种链式表,克服了上述的缺点,插入和删除操作均不会引起元素的移动:数据结构定义如下: public class ListNode { String val; ListNode next; // ... } 常见的链表有单向链表(也称之为chai…

Linked List Cycle II 题解 题目来源:https://leetcode.com/problems/linked-list-cycle-ii/description/ Description Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up: C…

Linked List Cycle 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/linked-list-cycle/description/ Description Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? Solution class Solution { publ…

Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 分析: 假设有环?遍历链表将无法走完,假设无环终会走到尾为NULL的位置 让一个指针每次走一个,一个指针每次走两个位置. 假设当中一个为NULL则无环. 假设相遇(必会相遇)了则有环. time,o(n),space,o(1) /** * Definition for sing…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 这个求单链表中的环的起始点是之前那个判断单链表中是否有环的延伸,可参见我之前的一篇文章 (http://www.cnblogs.com/grandyang/p/4137187.html). 还是要设…

Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 这道题是快慢指针的经典应用.只需要设两个指针,一个每次走一步的慢指针和一个每次走两步的快指针,如果链表里有环的话,两个指针最终肯定会相遇.实在是太巧妙了,要是我肯定想不出来.代码如下: C++ 解法: class Solution { public: bool hasCycle…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 解题思路,本题和上题十分类似,但是需要观察出一个规律,参考LeetCode:Linked List Cycle II JAVA实现如下: public ListNode detectCycle(Li…