1. Description: Notes: 2. Examples: 3.Solutions: 1/**2* Created by sheepcore on 2019-05-143* Definition for singly-linked list.4* class ListNode {5* int val;6* ListNode next;7* ListNode(int x) {8* val = x;9* next = null;10* }11* }12*/13publicbooleanhasCycle(ListNode head)...
/** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */publicclassSolution{publicbooleanhasCycle(ListNode head){if(head ==null)returnfalse;ListNodeslow=head;ListNodefast=head.next;while(fast...
利用LeetCode: 141. Linked List Cycle 题解 的快慢指针找到距离起点 n 个周期的节点(设慢指针移动 a+b 各节点, 则快指针移动 a+b+nT, 而快指针速度是慢指针的二倍,因此 2(a+b)=a+b+nT, 即 a+...
Return true if there is a cycle in the linked list. Otherwise, return false. 英文版地址 leetcode.com/problems/l 中文版描述 给你一个链表的头节点 head ,判断链表中是否有环。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 ...
LeetCode Top 100 Liked Questions 141. Linked List Cycle (Java版; Easy) 题目描述 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) ...
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]: # 如果是空链表,则直接返回 None if head is None: return...
141 Linked..判断链表 LinkList 是否带循环。Given a linked list, determine if it has a cycle in it.To represent a cycle in t
Explanation: There is a cycle in the linked list, where tail connects to the second node. image.png 二、解决思路 方法一:使用HashMap存储遍历链表,并判重,O(n) 方法二:使用一快一慢指针检查是否相等,O(n) 三、算法实现 publicstatic booleanisCycle(Node head){if(head==null)returnfalse;boolean fla...
public class Solution { /** * @param head: The first node of linked list. * @return: The node where the cycle begins. * if there is no cycle, return null */ public ListNode detectCycle(ListNode head) { if(head == null || head.next == null) ...
Given a linked list, determine if it has a cycle in it. 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。 [java]view plaincopy /** * Definition for singly-linked list. ...