Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 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.…

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? 141. Linked List Cycle 的拓展,这题要返回环开始的节点,如果没有环返回null. 解法:双指针,还是用快慢两个指针,相遇时记下节点.参考:willduan的博客 Java: pu…

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…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 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.…

一.题目大意 https://leetcode.cn/problems/linked-list-cycle-ii/ 给定一个链表的头节点  head ,返回链表开始入环的第一个节点. 如果链表无环,则返回 null. 如果链表中有某个节点,可以通过连续跟踪 next 指针再次到达,则链表中存在环. 为了表示给定链表中的环,评测系统内部使用整数 pos 来表示链表尾连接到链表中的位置(索引从 0 开始).如果 pos 是 -1,则在该链表中没有环.注意:pos 不作为参数进行传递,仅仅是为了标识链…

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, 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? 参考http://www.cnblogs.com/hiddenfox/p/3408931.html 方法: 第一次相遇时slo…

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? 141题的延伸,求出循环点. 可以用数学方法证明出slow与find相遇的位置一定是所求的点. /** * Definitio…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 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. I…

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…

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…

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…

142. Linked List Cycle II[easy] 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? 解法一: /** * Definition for singl…

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…

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…

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:由[Leetcode]Linked List Cycle可知.利用一快一慢两个指针可以推断出链表是否存在环路. 如果两个指针相遇之前slow走了s步,则fast走了2s步.而且fast已经在长…

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…

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…

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…

这是LeetCode里的第142道题. 题目要求: 给定一个链表,返回链表开始入环的第一个节点. 如果链表无环,则返回 null. 说明:不允许修改给定的链表. 进阶:你是否可以不用额外空间解决此题? 起初我在做这道题的时候,以为挺简单的,以为循环链表都是已头节点为循环头,结果... ~~~~(>_<)~~~~ 没考虑到链中任一个节点都可能是循环头的头节点. 一开始比较贪心,就只设置的一个指针p来判断是否循环,结果思考不充分,没考虑到第二种特殊的情况,导致错了很多次. 然后经过多次测试后终于成…

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, 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? 头结点到cycle begins的点 距离是A, cycle begins的点 到快慢结点相遇的 点的距离是B A+B+N =…

给一个链表,返回链表开始入环的第一个节点. 如果链表无环,则返回 null.说明:不应修改给定的链表.补充:你是否可以不用额外空间解决此题?详见:https://leetcode.com/problems/linked-list-cycle-ii/description/ Java实现: /** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x)…

题目意思:如果有环,返回入口结点 思路:先判断有没环,再计算环的结点数,然后p1指向头,p2往后移结点次数,p1.p2相遇为入口结点 ps:还是利用指针间距这个思路 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: Li…

题目: 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? 给定一个链表的头指针,问你能不能只用常数的空间快速判断一个链表是不是有环,如果有环,返回环的起始位置. 代码: 不能…

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? 题意: 给定一个链表,找到环起始的位置.如果环不存在,返回NULL. 分析: (1)首先要判断该链表是否有环.如果没有环,那么返回NULL. (2)其次,当已知环存在后,寻找环起始的位置. 思路: (…

公司和学校事情比较多,隔了好几天没刷题,今天继续刷起来. 题目: 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? 代码:oj 测试通过 Runtime: 596 ms # Definition for singly-linked list. # c…

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: 做过,就当复习了. 先用快慢指针判断相交,关键是环开始点的获取. 用上图说明一下,设非环的部分长度为a(包括环的入口点), 环的长度为b(包括环的入口点). 快慢指针相交的位置为绿色的点,距离…

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? 思路:设head距离循环开始点k,循环开始点距离fast和slow第一次相遇点x,slow还要走y到达循环开始点.则有:x+…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 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.…