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. 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. 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. 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 =…

这道题是LeetCode里的第141道题. 题目要求: 给定一个链表,判断链表中是否有环. 进阶: 你能否不使用额外空间解决此题? 简单题,但是还是得学一下这道题的做法,这道题是用双指针一个fast,一个slow.fast每一步前进两个节点,slow前进一个节点.判断fast和slow是否相等或者为空就行了. 贴个代码: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next…

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…

给定一个链表,判断链表中否有环.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/description/ Java实现: /** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } *…

题意:判断链表是否有环. 分析: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) { ListNode* fast = head; ListNode…

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, 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…