LeetCode | Linked List Cycle II

来源：互联网发布：云计算架构师要求编辑：程序博客网时间：2024/06/06 09:26

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?

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */

//思路：
首先利用快慢双指针向后遍历（slow每次向后一格，fast每次向后两格），当两个指针重合时，则list中一定存在cycle
然后让fast_pointer停在两指针相遇的位置，而让slow_pointer回到head，让两指针同步的每次向后遍历一格，则两指针相遇的node即为cycle begin的node（有点像单纯的智力题。。。）

public class Solution {    public ListNode detectCycle(ListNode head) {        if(head==null || head.next==null) return null;                ListNode slow_pointer = head;        ListNode fast_pointer = head;                boolean hasCycle = false;        while(fast_pointer!=null && fast_pointer.next!=null){  //判断list中是否有cycle的存在，思路与昨天的题一样            slow_pointer = slow_pointer.next;            fast_pointer = fast_pointer.next.next;            if(slow_pointer == fast_pointer){                hasCycle = true;                break;                //注意：此处要有break，否则两个pointer会在cycle中一直遍历下去，无限循环            }        }                if(hasCycle){                    slow_pointer = head;            while(true){                if(slow_pointer == fast_pointer) return slow_pointer;                slow_pointer = slow_pointer.next;                fast_pointer = fast_pointer.next;          //if(slow_pointer == fast_pointer) return slow_pointer;  注：判断的语句不能放在后边，不然当list中tail.next=head时会出错            }        }                return null;    }}

0 0