142. Circular linked list||

The question ： Given a linked list , Return to the first node of the link where the list begins to enter .  If the list has no links , Then return to  null.

To represent a ring in a given list , Use integers 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 .

Advanced challenges ： It is not allowed to modify the given linked list .

Method ： Speed pointer （ Double pointer ）

public class Huanxinglianbiao2 {
    public ListNode detectCycle(ListNode head){
        // Set speed pointer 
        ListNode fast = head;
        ListNode slow = head;
        while(fast!=null && fast.next!=null){
            // Let's go two steps 
            // Slow pointer one step 
            fast = fast.next.next;
            slow = slow.next;

            // If fast == slow Confirm that there is a ring 
            // Find the meeting point of the fast and slow pointers , Meet again. Come back to head, The slow pointer remains stationary 
            // They move forward at the same speed , When they meet again , It's the entry point of the circular entrance 
            if (fast == slow){
                fast = head;
                while (fast!=slow){
                    fast = fast.next;
                    slow = slow.next;
                }
                // Back to the entrance of the ring 
                return fast;
            }
        }
        // End of traversing the linked list , Ring not found , It returns null ;
        return null;
    }
}