Mathematics to solve the problem - circular linked list

Circular linked list

Things:

1. First traverse the linked list through the fast and slow double pointers until the fast and slow meet and stop (The speed of fast must be twice the speed of slow, not three or four times... Imagine if the ring has only two knotsPoint, the fast must be advanced in the ring, if the slow in the backward ring is just staggered from the fast, the slow takes three times, and the fast takes two steps, and they will never meet!)
2. Let one pointer be traversed from the head node, and the other pointer will be traversed from the point where the speed and slow pointers met last time, The traversal speed of the two pointers is the same and they will meet at the node entrance of the loop, I can't think of it?The explanation is as follows:

The code is as follows:

public class Solution {public ListNode detectCycle(ListNode head) {//fast pointer must be twice as fast as slow pointerListNode fast = head;ListNode slow = head;while(fast != null && fast.next != null){fast = fast.next.next;slow = slow.next;if(fast == slow){break;}}if(fast == null || fast.next == null){return null;}slow = head;while(slow != fast){slow = slow.next;fast = fast.next;}return slow;}}