程序媛詹妮弗
终身学习

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.

 

 

 

题意:

找出循环链表的入口

 

思路

Two Pointers

脑筋急转弯题。没有深究的价值。

 

code

 1 /**
 2  * Definition for singly-linked list.
 3  * class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode detectCycle(ListNode head) {
14         ListNode slow = head, fast = head;
15         while (fast != null && fast.next != null) {
16             slow = slow.next;
17             fast = fast.next.next;
18             if (slow == fast) {
19                 ListNode slow2 = head;
20                 while (slow2 != slow) {
21                     slow2 = slow2.next;
22                     slow = slow.next;
23                 }
24                 return slow;
25             }
26         }
27         return null;
28     }
29 }

 

posted on 2019-05-16 07:06  程序媛詹妮弗  阅读(151)  评论(0编辑  收藏  举报