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 con
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= p->next;//p每次前进一步if(!q->ne...
}publicstaticvoidmain(String[] args){LinkedListCyclelinkedListCycle=newLinkedListCycle();LinkedNodelist=linkedListCycle.createList(newint[]{1,2,3,4,5}); System.out.println(linkedListCycle.hasCycle(list) +" == false"); linkedListCycle.makeCycle(list,2); System.out.println(linkedListCycle.hasCycle(...
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 the linked list. 翻译过来就是找到链表中的环,并返回环起点的节点。
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 ...
https://leetcode.com/problems/linked-list-cycle-ii/ 题目: null. Note: Follow up: Can you solve it without using extra space? 思路: 首先确定循环是否存在,若存在,根据 循环结点个数/结点相对移动次数 就会相遇的规律 得到循环结点个数,再从头开始遍历,相对移动速度为结点个数,此时两指针第一次相遇的位...
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;/...