slow 和 fast 节点分别初始化为节点 1 和 2,假设快慢指针第一次相遇的节点为 0 对应于环中的第 i 个节点 Ci, 那么此时慢指针正好走了 n−r−1+i 步,快指针则走了 2⋅(n−r−1+i) 步,且存在式1: n−r−1+i+1=l⋅r. (之所以在 i 后面加 1 是因为快指针初始化时多走了一步) 快慢指针第一次相遇时慢
Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull. Example Given-21->10->4->5, tail connects to node index 1,return10 Challenge Follow up: Can you solve it without using extra space? /// /*** Definition for ListNode. * public class Lis...
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。
public ListNode detectCycle(ListNode head) { HashSet<ListNode> set = new HashSet<>(); while (head != null) { set.add(head); head = head.next; if (set.contains(head)) { return head; } } return null; } 解法二 快慢指针 还是之前的思想, 学数据结构课程的时候,应该都用过这个方法,很...
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: Can you solve it without using extra space? 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。
Explanation: 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 list, where tail connects to the first node. Example 3: Input: head...
Example 2: Input:head=[1,2],pos=0Output:tail connects to node index0Explanation:There is a cycle in the linked list,where tail connects to the first node. 1. 2. 3. Example 3: Input:head=[1],pos=-1Output:no cycle Explanation:There is no cycle in the linked list. ...
leetcode 之Linked List Cycle(24) 两个思路,一是用哈希表记录每个结点是还被访问过;二是定义两个快、慢指针,如果存在环的话,两个指针必定会在某位结点相遇。 boollinkListNode(ListNode *head) { ListNode*fast=head, *slow=head;while(fast && fast->next)...
Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 译文:给定一个链表,如果链表中存在环,则返回到链表中环的起始节点,如果没有环,返回null。 知识点 Linked List Cycle I —— 通过快慢两个指针来确定链表上是否存在环 快慢指针的相遇点到环入口点的距离 ...
if(slow==fast)returnfast;[1,2]tail connects to node index0 第一次检查cycle时候 fast == slow == 1 接下来循环入口如果不判断,之后两个会在[2]判断相等并返回[2]作为cycle的头结点,明显是不对的。 structListNode*detectCycle(structListNode*head){if(head==NULL)returnNULL;structListNode*slow,*fast...