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Editor to share with you how to solve the problem of leetcode linked list loop detection, I believe that most people do not know much about it, so share this article for your reference, I hope you can learn a lot after reading this article, let's go to know it!

Given a linked list, if it is a linked list, implement an algorithm to return the beginning node of the loop. There is a definition of a linked list: the next element of a node in the linked list points to the node that appeared before it, indicating that there is a loop in the linked list. Example 1: input: head = [3jin2j0jimi 4], pos = 1 output: tail connects to node index 1 explanation: there is a ring in the linked list, its tail is connected to the second node. Example 2: input: head = [1Power2], pos = 0 output: tail connects to node index 0 explanation: there is a ring in the linked list whose tail is connected to the first node. Example 3: input: head = [1], pos =-1 output: no cycle explanation: there are no rings in the linked list. Advanced: can you solve this problem without extra space? Source: LeetCode link: https://leetcode-cn.com/problems/linked-list-cycle-lcci copyright belongs to the collar buckle network. For commercial reprint, please contact official authorization. For non-commercial reprint, please indicate the source. Answer the question / Definition for singly-linked list. * class ListNode {* int val; * ListNode next; * ListNode (int x) {* val = x; * next = null; *} * * / public class Solution {public ListNode detectCycle (ListNode head) {ListNode slow = head; ListNode fast = head; while (fast! = null & & fast.next! = null) {slow = slow.next Fast = fast.next.next; if (slow = = fast) {break;}} if (fast = = null | | fast.next = = null) {return null;} while (head! = fast) {head = head.next; fast = fast.next;} return head }} above are all the contents of this article entitled "how to solve the loop detection problem of leetcode linked list". Thank you for reading! I believe we all have a certain understanding, hope to share the content to help you, if you want to learn more knowledge, welcome to follow the industry information channel!

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