LeetCode-in-All

142. Linked List Cycle II

Medium

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

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

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.

Constraints:

Follow up: Can you solve it using O(1) (i.e. constant) memory?

Solution

import com_github_leetcode.ListNode

/*
 * Definition for singly-linked list.
 * class ListNode(var _x: Int = 0) {
 *   var next: ListNode = null
 *   var x: Int = _x
 * }
 */
object Solution {
    def detectCycle(head: ListNode): ListNode = {
        def getIntersect: Option[ListNode] = {
            var slow = head
            var fast = head

            while (fast != null && fast.next != null) {
                slow = slow.next
                fast = fast.next.next
                if (fast == slow) return Some(slow)
            }
            None
        }

        getIntersect match {
            case None => null
            case node => {
                var ptr1 = head
                var ptr2 = node.get

                while (ptr1 != ptr2) {
                    ptr1 = ptr1.next
                    ptr2 = ptr2.next
                }
                ptr1
            }
        }
    }
}