142. Linked List Cycle II

Description

Given 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. Note that pos is not passed as a parameter.

Notice that you should not modify the linked list.

Constraints

• The number of the nodes in the list is in the range [0, 104].

• -105 <= Node.val <= 105

• pos is -1 or a valid index in the linked-list.

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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.

Solutions

ListNode

// Definition for singly-linked list.
class ListNode {
    int val;
    ListNode next;
    
    ListNode(int x) {
        val = x;
        next = null;
    }
}

Solution 1

/**
 * Time complexity : O(N)
 * Space complexity : O(N)
 */

public class Solution {
    public ListNode detectCycle(ListNode head) {
        Set<ListNode> visited = new HashSet<ListNode>();

        ListNode node = head;
        while (node != null) {
            if (visited.contains(node)) {
                return node;
            }
            visited.add(node);
            node = node.next;
        }

        return null;
    }
}

Solution 2

/**
 * Time complexity : O(N)
 * Space complexity : O(1)
 */
 
public class Solution {
    public ListNode detectCycle(ListNode head) {
        if(head == null) return null;
        
        ListNode slowptr = head;
        ListNode fastptr = head;
        
        do {
            if(fastptr == null || fastptr.next == null) {
                return null;
            }
            slowptr = slowptr.next;
            fastptr = fastptr.next.next;
        } while(slowptr != fastptr);
        
        ListNode startptr = head;
        while(startptr != slowptr) {
            startptr = startptr.next;
            slowptr = slowptr.next;
        }
        return startptr;
    }
}

Follow up

• Can you solve it using O(1) (i.e. constant) memory?