142. Linked List Cycle II(重排链表)

最新推荐文章于 2021-07-13 22:43:30 发布

原创最新推荐文章于 2021-07-13 22:43:30 发布 · 192 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Linked List Cycle II #重排链表 #双指针

Leetcode 专栏收录该内容

227 篇文章

订阅专栏

本文介绍了一种使用快慢指针技术解决链表中寻找环的入口节点问题的方法。通过实验证明，当快慢指针在环内首次相遇后，将快指针移回头结点并同时以相同速度前进，再次相遇的节点即为环的入口。此方法无需额外空间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：https://leetcode.com/problems/linked-list-cycle-ii/

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 no cycle in the linked list.

Note: Do not modify the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

Follow up:
Can you solve it without using extra space?

思路：快慢指针

具体的严密证明我没有去做，我只是找几个带环的例子进行了实际的测试，发现，当fast

和slow都从head出发，fast以2倍速，slow以一倍速，当两个节点相遇，然后fast再改为一倍速，

slow仍是一倍速，那么下次fast和slow相遇的节点就是环的入口节点。

AC 0ms Java：

/**
 * 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) {
        if(head==null||head.next==null)
            return null;
        ListNode fast=head;
        ListNode slow=head;
        boolean isCircle=false;
        while(fast!=null&&fast.next!=null){
            fast=fast.next.next;
            slow=slow.next;
            if(fast==slow){
                isCircle=true;
                break;
            }
        }
        if(!isCircle)
            return null;
        fast=head;
        while(isCircle&&fast!=slow){
            fast=fast.next;
            slow=slow.next;
        }
        return fast;
    }
}