Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? Solution: solve the problem with one and two steps no cycle case: the faster pointer(two step) reaches the null first cycle : slower == fast…

题目: 141.Given a linked list, determine if it has a cycle in it. 142.Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 思路: 带环链表如图所示.设置一个快指针和一个慢指针,快指针一次走两步,慢指针一次走一步.快指针先进入环,慢指针后进入环.在进入环后,可以理解为快指针追赶慢指针,由于两个指…

判断链表有环,环的入口结点,环的长度 1.判断有环: 快慢指针,一个移动一次,一个移动两次 2.环的入口结点: 相遇的结点不一定是入口节点,所以y表示入口节点到相遇节点的距离 n是环的个数 w + n + y = 2 (w + y) 经过化简,我们可以得到:w  = n - y; https://www.cnblogs.com/zhuzhenwei918/p/7491892.html 3.环的长度: 从入口结点或者相遇的结点移动到下一次再碰到这个结点计数 https://blog.csdn.ne…

141. 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? 利用快慢指针,如果相遇则证明有环 注意边界条件: 如果只有一个node. public class Solution { public boolean hasCycle(ListNode head) { if(head==null |…

引入 快慢指针经常用于链表(linked list)中环(Cycle)相关的问题.LeetCode中对应题目分别是: 141. Linked List Cycle 判断linked list中是否有环 142. Linked List Cycle II 找到环的起始节点(entry node)位置. 简介 快指针(fast pointer)和慢指针(slow pointer)都从链表的head出发. slow pointer每次移动一格,而快指针每次移动两格. 如果快慢指针能相遇,则证明链表中有…

141. Linked List Cycle[easy] 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: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x…

Question 141. Linked List Cycle Solution 题目大意:给一个链表,判断是否存在循环,最好不要使用额外空间 思路:定义一个假节点fakeNext,遍历这个链表,判断该节点的next与假节点是否相等,如果不等为该节点的next赋值成fakeNext Java实现: public boolean hasCycle(ListNode head) { // check if head null if (head == null) return false; ListN…

Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in…

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…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 给定一个链表,判断是否有环存在.Follow up: 不使用额外空间. 解法:双指针,一个慢指针每次走1步,一个快指针每次走2步的,如果有环的话,两个指针肯定会相遇. Java: public class Solution { public boolean hasCycle(Li…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 题意: 判断一个链表是否有环. 解决方案: 双指针,快指针每次走两步,慢指针每次走一步, 如果有环,快指针和慢指针会在环内相遇,fast == slow,这时候返回true. 如果没有环,返回false. /** * Definition for singly-linked li…

Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 解题思路: 由于不让用extra space,所以用一个快指针和一个慢指针,快指针一次移动两步,慢指针一次移动一步,只要快指针赶上慢指针证明纯在loop,JAVA实现如下: public boolean hasCycle(ListNode head) { ListNode fa…

Given a linked list, determine if it has a cycle in it. 代码如下: /** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public boolean hasC…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 判断是否成环 1.利用set,很简单,但是题目中说不要用额外的空间. /** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x)…

题目描述: Given a linked list, determine if it has a cycle in it. 解题思路: 快的指针和慢的指针 代码如下: # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def hasCycle(self, h…

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…

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…

Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer poswhich represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in t…

Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in…

题目: Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 题意: 给定一个链表,判断是否有环 思路: 快慢指针 若有环,则快慢指针一定会在某个节点相遇(此处省略证明) 代码: public class Solution { public boolean hasCycle(ListNode head) { ListNode fast =…

Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 思路:采用“快慢指针”查检查链表是否含有环.让一个指针一次走一步,另一个一次走两步,如果链表中含有环,快的指针会再次和慢的指针相遇. class Solution { public: bool hasCycle(ListNode *head) { ListNode* slow…

Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 题目标签:Linked List 题目给了我们一个 Linked List,让我们判断它是否循环. 利用快,慢指针,快指针一次走2步,慢指针一次走1步,如果循环,快慢指针一定会相遇. Java Solution: Runtime beats 98.15% 完成日期:06/09/2…

Problem describe:https://leetcode.com/problems/linked-list-cycle/ Given a linked list, determine if it has a cycle in it. To represent a cycle -indexed) , then there is no cycle in the linked list. Example : Input: head = [,,,-], pos = Output: true E…

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…

#-*- coding: UTF-8 -*- #Method:快慢指针法,建立虚表头,快指针走两步,慢指针走一步,若存在环,则快指针会追上慢指针# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def hasCycle(s…

题目意思:链表有环,返回true,否则返回false 思路:两个指针,一快一慢,能相遇则有环,为空了没环 ps:很多链表的题目:都可以采用这种思路 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool hasCyc…

非常简单的题:判断链表有没有环(用快慢指针) /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool hasCycle(ListNode *head) { if (head == NULL) return false…