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 = 1Output: tail connects to node index 1Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

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

Example 3:

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

 

Follow up:

Can you solve it without using extra space?

 

Floyd's cycle detection algorithm, aka Tortoise and Hare Algorithm

ref: 

time: O(n), space: O(1)

/** * 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 slow = head, fast = head, start = head;        while(fast != null && fast.next != null) {            fast = fast.next.next;            slow = slow.next;            if(fast == slow) {                while(slow != start) {                    slow = slow.next;                    start = start.next;                }                return start;            }        }        return null;    }}