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?
解题思路：双指针， 一个指针一次走一步，一个指针一次走两步， 如果有环，定相遇。然后将快指针移到头部，再一次走一步，当两只真再次相遇的时候就是指针的头
/** * 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) return null; ListNode fast = head; ListNode slow = head; while(fast.next != null && fast.next.next != null){ fast = fast.next.next; slow = slow.next; if(fast == slow){ break; } } if(fast.next == null || fast.next.next == null) return null; fast = head; while(fast != slow){ fast = fast.next; slow = slow.next; } return fast; } }
原文：http://www.cnblogs.com/RazerLu/p/3554151.html