Linked List Cycle II

Given a linked list, return the node where the cycle begins.

If there is no cycle, return null.

Have you met this question in a real interview? 

Yes

Example

Given -21->10->4->5, tail connects to node index 1，return 10


来源： http://www.lintcode.com/en/problem/linked-list-cycle-ii/#

分析

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/**  * Definition for ListNode.  * public class ListNode {  *     int val;  *     ListNode next;  *     ListNode(int val) {  *         this.val = val;  *         this.next = null;  *     }  * }  */  public class Solution {     /**      * @param head: The first node of linked list.      * @return: The node where the cycle begins.       *           if there is no cycle, return null      */     public ListNode detectCycle(ListNode head) {           // write your code here         ListNode slow = head, fast = head;         ListNode Pos = null;         while(fast != null && fast.next != null && fast.next.next != null){             slow = slow.next;             fast = fast.next.next;             if(fast == slow){                 Pos = slow;                 break;             }         }         if(Pos == null) return null;                   while(head != Pos){             head = head.next;             Pos = Pos.next;         }         return Pos;     } }

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