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置就是循环开始。 算法: public ListNode detectCycle(ListNode head) { ListNode q = head, p = null; boolean flag = false; // 确定循环...
此时将其中一个指针重新指向链头,然后两个指针每次移动一步,直到再次相遇; 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 = h...
Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull. To represent a cycle in the given linked list, we use an integerposwhich represents the position (0-indexed) in the linked list where tail connects to. Ifposis-1, then there is no cycle i...
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 ...
Explanation: There is a cycle in the linked list, where tail connects to the first node. 1. 2. 3. Example 3: AI检测代码解析 Input: head = [1], pos = -1 Output: no cycle Explanation: There is no cycle in the linked list. ...
next = pre; // 由于已是最后一次插入,所以无需再移动尾结点 } // 返回结果链表的头结点 head_pre.next } } 题目链接: Linked List Cycle : leetcode.com/problems/l 环形链表: leetcode-cn.com/problem LeetCode 日更第 53 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
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
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* @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) return null; ...