LinkedListCycle.java

// Source : https://leetcode.com/problems/linked-list-cycle
// Author : cornprincess
// Date   : 2020-04-21

/***************************************************************************************************** 
 *
 * Given a linked list, determine if it has a cycle in it.
 * 
 * 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.
 * 
 * Example 1:
 * 
 * Input: head = [3,2,0,-4], pos = 1
 * Output: true
 * 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: true
 * Explanation: There is a cycle in the linked list, where tail connects to the first node.
 * 
 * Example 3:
 * 
 * Input: head = [1], pos = -1
 * Output: false
 * Explanation: There is no cycle in the linked list.
 * 
 * Follow up:
 * 
 * Can you solve it using O(1) (i.e. constant) memory?
 ******************************************************************************************************/

package LinkedListCycle;

import java.util.HashSet;
import java.util.Set;

public class LinkedListCycle {

    //Time complexity: O(n)
    //Space complexity: O(n)
    public boolean hashSet(ListNode head) {
        Set<ListNode> nodesSeen = new HashSet<>();

        while (head != null) {
            if (nodesSeen.contains(head)) {
                return true;
            } else {
                nodesSeen.add(head);
            }
            head = head.next;
        }
        return false;
    }

    //Time complexity: O(n)
    //Space complexity: O(1)
    public boolean twoPoint(ListNode head) {
        ListNode slow = head;
        ListNode fast = head.next;

        while (slow != fast) {
            if (fast == null || fast.next == null) {
                return false;
            }
            slow = slow.next;
            fast = fast.next.next;
        }
        return true;
    }
}