Linked List Cycle II

LeetCode 142 | Difficulty: Medium

Medium

Problem Description

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:

- The number of the nodes in the list is in the range `[0, 10^4]`.

- `-10^5 <= Node.val <= 10^5`

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

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

Topics: Hash Table, Linked List, Two Pointers

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Approach

Hash Map

Use a hash map for O(1) average lookups. Store seen values, frequencies, or indices. The key question: what should I store as key, and what as value?

When to use

Need fast lookups, counting frequencies, finding complements/pairs.

Linked List

Use pointer manipulation. Common techniques: dummy head node to simplify edge cases, fast/slow pointers for cycle detection and middle finding, prev/curr/next pattern for reversal.

When to use

In-place list manipulation, cycle detection, merging lists, finding the k-th node.

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Solutions

Solution 1: C# (Best: 162 ms)

Metric	Value
Runtime	162 ms
Memory	N/A
Date	2017-09-23

Solution

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

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Complexity Analysis

Approach	Time	Space
Two Pointers	$O(n)$	$O(1)$
Hash Map	$O(n)$	$O(n)$
Linked List	$O(n)$	$O(1)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Hash map gives O(1) lookup — think about what to use as key vs value.
• Draw the pointer changes before coding. A dummy head node simplifies edge cases.