Given a linked list, return the node where the cycle begins. If there is no cycle, return null
.
Follow up:
Can you solve it without using extra space?
Solution:
快慢指针。
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 if(head==null||head.next==null) 15 return null; 16 ListNode fast=head; 17 ListNode slow=head; 18 while(fast!=null&&fast.next!=null){ 19 fast=fast.next.next; 20 slow=slow.next; 21 if(slow==fast){ 22 fast=head; 23 while(fast!=slow){ 24 fast=fast.next; 25 slow=slow.next; 26 } 27 return slow; 28 } 29 } 30 return null; 31 } 32 }