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
Input:head = [3,2,0,-4], pos = 1Output:tail connects to node index 1Explanation:There is a cycle in the linked list, where tail connects to the second node. Example 2: Input:head = [1,2], pos = 0Output:tail connects to node index 0Explanation:There is a cycle in the linked ...
这就清晰了,X和K的关系是基于Y互补的。等于说,两个指针相遇以后,再往下走X步就回到Cycle的起点了。这就可以有O(n)的实现了。 from : http://fisherlei.blogspot.tw/2013/11/leetcode-linked-list-cycle-ii-solution.html Code : /** * Definition for singly-linked list. * struct ListNode { * int ...
next = ListNode(list[i]) cur = cur.next return head ## 头元素指代一个链表 这里暂时还没想明白怎么初始化带有环的链表,所以这次就先直接在LeetCode中提交 solution。 class Solution: def hasCycle(self, head:ListNode) -> bool: seen = set() ## 空集合遍历收集元素 while head: ## 存在头结点 ...
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...
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...
leetcode -- Linked List Cycle -- 重点 一开始我错误的code: class Solution(object): def hasCycle(self, head): """ :type head: ListNode :rtype: bool """ if not head : return False slow, fast = head, head while fast and fast.next and fast != slow:#这里这个条件fast!=slow,使得在...
Explanation: There is no cycle in the linked list.我看到就蒙了 感觉我工作上头基本没有用过linklist这种数据结构。 都是List 或者Dictionary 或者array那么这个LinkList如果是一个类的话 类的实例(引用类型)可以直接比较吗?我就写了如下的solution 虽然accepted了 但是 faster than0.86% 感觉这题还得重做。我...
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...