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Linked List Cycle

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?

Implementation 1 : Time : O(n), Space : O(n)

/**
 * 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)
            return false;
        Set<ListNode> set = new HashSet<>();
        ListNode current = head;
        while(current != null) {
            if(set.contains(current))
                return true;
            set.add(current);
            current = current.next;
        }
        return false;
    }
}

Implementation 2: (Fast & Slow will eventually meet 🤝 )

/**
 * 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 there is no node or just one node in the linked list
        if (head == null || head.next == null) {
            return false;
        }
        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;
    }
}

Implementation 2: Just little different (Fast & Slow will eventually meet 🤝 )

/**
 * 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)
            return false;
        
        ListNode slow = head;
        ListNode fast = head;
        
        while(fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast)
                return true;
            
        }
        return false;
    }
}

References :

  1. https://leetcode.com/articles/linked-list-cycle
  2. https://www.youtube.com/watch?v=6OrZ4wAy4uE