142. Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

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

 

Similar: 141. Linked List Cycle

 

 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 fast = head, slow = head;
15         while (fast != null) {
16             fast = fast.next;
17             if (fast == null) return null;
18             fast = fast.next;
19             slow = slow.next;
20             if (fast == slow) break;
21         } 
22         if (fast == null) return null;
23         slow = head;
24         while (slow != fast) {
25             slow = slow.next;
26             fast = fast.next;
27         }
28         return slow;
29     }
30 }

 

posted @ 2016-04-20 12:04  Joyce-Lee  阅读(173)  评论(0)  编辑  收藏  举报