Input:head = [1,2], pos = 0Output:tail connects to node index 0Explanation:There is a cycle in the linked list, where tail connects to the first node. Example 3: Input:head = [1], pos = -1Output:no cycleExplana
方法是利用两个指针从头開始,指针p1一次走一步,指针p2一次走两步,假设有环则两指针必然有重逢之时(这是Linked List Cycle里用到的)。 然后就是怎样求出环的起始节点。 能够这么假定,从链的起点到环的起点,这段距离称为a。环的长度称为c,第一次相遇位置距环的起点距离为p。首先p1被p2追上时它一定没有走完...
LeetCode: Linked List Cycle 解题报告 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? SOLUTION 1: 经典快慢指针问题。如果存在环,fast, slow必然会相遇。就像2个速度不一样的人在环形跑道赛跑,总有一个时间他们会相...
来自专栏 · LeetCode Description 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. If...
Leetcode: Linked List Cycle 题目: Given a linked list, determine if it has a cycle in it. 思路分析: 利用快慢指针slow,fast。 slow指针每次走一步,fast指针每次走两步,倘若存在环,则slow和fast必定在某一时刻相遇。 C++参考代码: /** * Definition for singly-linked list....
# 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...
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...
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
Problem link: https://leetcode.com/problems/linked-list-cycle-ii/有任何错误欢迎指出,有任何问题欢迎留言,谢谢观看, 视频播放量 94、弹幕量 1、点赞数 1、投硬币枚数 2、收藏人数 1、转发人数 1, 视频作者 dddeng12, 作者简介 ,相关视频:51. N 皇后 (N-queens),重
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...