leetcode 142. Circular linked list II

difficulty ： secondary
The frequency of ：114
subject ： Given the head node of a linked list head , Return to the first node of the link where the list begins to enter . If the list has no links , Then return to null.

If there is a node in the linked list , It can be done by continuously tracking next The pointer reaches again , Then there is a ring in the linked list . To represent a ring in a given list , The evaluation system uses an integer pos To indicate where the end of the list is connected to the list （ Index from 0 Start ）. If pos yes -1, There are no links in the list . Be careful ：pos Not passed as an argument , Just to identify the actual situation of the linked list .


Their thinking ： Speed pointer
Be careful ：

• law ： The length from the node where the fast and slow pointers meet to the starting node of the ring =head The length to the ring start node 【 Use the slow pointer to walk , A from head Start , A place where the fast and slow pointers meet , Finally, the index is 0 The nodes meet 】
• Practice the circular linked list 141 while Don't forget in your judgment fast.next!=null
  Code ：

/** * 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) {
    
        ListNode fast=head,slow=head;
        //fast.next!=null Can't forget   If you take two steps, you may encounter the next step and there is no way to go 
        while(fast!=null&&fast.next!=null){
     
            fast=fast.next.next;
            slow=slow.next;
            if(fast==slow) {
    
                fast=head;
                while(fast!=slow){
    
                    fast=fast.next;
                    slow=slow.next;
                }
                return slow;
            } 
        }
        return null;
    }
}