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?
这道题其实是一个数学题,先检测出是不是有环,在有环的情况下slow再走x步(x是环外的节点个数)就可以到达环的开始点。
/** * 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; while(fast != null && fast.next !=null){ fast = fast.next.next; slow = slow.next; if(fast == slow){ break; } } if(fast != slow){ return null; } fast = head; while(fast != slow){ fast = fast.next; slow = slow.next; } return fast; } }
原文:http://www.cnblogs.com/incrediblechangshuo/p/4348874.html