44.Linked List Cycle II（环的入口节点）

Posted 2022-02-09 yjxyy

Level：

??Medium

题目描述：

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

Follow up:
Can you solve it without using extra space?

思路分析：

??如果存在环，设置一个快指针，一个慢指针，那么快指针一定会追上慢指针相遇，此时相遇的节点一定在环内，这时可以求出环内节点的数目，然后设置一个前指针和后指针初始值都为head，让前指针先走n次，然后前后指针一起走，如果相等时，则该节点就为环入口节点

代码：

public class Solution
    public ListNode detectCycle(ListNode head)
        ListNode meetNode=meetNoding(head);
        if(meetNode==null)
            return null;
        ListNode pNode=meetNode;
        int count=1;
        while(pNode.next!=meetNode)
            count++;
            pNode=pNode.next;
        
        ListNode slow=head;
        ListNode fast=head;
        for(int i=0;i<count;i++)
            slow=slow.next;
        
        while(fast!=slow)
            fast=fast.next;
            slow=slow.next;
        
        return fast;
    
    //求相遇的节点
    public ListNode meetNoding(ListNode head)
        if(head==null)
            return null;
        if(head.next==null)
            return null;
        ListNode slow=head.next;
        ListNode fast=slow.next;
        while(slow!=null&&fast!=null)
            if(slow==fast)
                return fast;
            slow=slow.next;
            fast=fast.next;
           if(fast!=null&&fast.next!=null)
               fast=fast.next; //快指针一次走两步
        
        return null; //未能相遇则不存在环