141. Linked List Cycle

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

Solution:

This is a classic Floyd Loop Detection problem.

We keep two pointers, one slow and one faster.

Each time the slow pointer move one step forward, while the faster pointer move two steps forward.

If there is a cycle, they will eventually meet at some point.



Code:

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) {
            return false;
        }
        
        ListNode fast = head.next;
        ListNode slow = head;
        while (slow != fast) {
            if (fast == null || fast.next == null) {
                return false;
            }
            fast = fast.next.next;
            slow = slow.next;
        }
        
        return true;
    }
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) {
            return false;
        }
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (slow == fast) {
                return true;
            }
        }
        
        return false;
    }
}