此时,另一个指针沿着环移动到环头需要步数:k*b - c = a;(k >= 1) 所以,两个指针第二次重合必定在环的开头位置。 ListNode *LeetCode::detectCycle(ListNode *head){if(!head)returnnullptr;boolhasCycle =false; ListNode*p = head, *q =head;while(q){ p
Example 1: 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...
运行时间0ms,这里要注意刚开始的判断和while循环结束条件,因为hare跑的比tortoise快,所以对于循环结束的判断只要判断hare就行了。 2、Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull. Note: Do not modify the linked list. Follow up...
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 ...
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 pos is -1, then there is ...
LeetCode 142:环形链表 II Linked List Cycle II 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。 为了表示给定链表中的环,我们使用整数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。
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....
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...
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
publicclassSolution{publicListNodedetectCycle(ListNode head){booleanisfast=true;ListNodeslow= head, fast = head;while(fast !=null&& fast.next !=null){ slow = slow.next; fast = isfast ? fast.next.next : fast.next;if(slow == fast){if(isfast){// 第一次相遇 isfast =false;/...