Linked List Cycle

LeetCode 141 | Difficulty: Easy

Easy

Problem Description

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

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.

Return true if there is a cycle in the linked list. Otherwise, return false.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.

Example 3:

Input: head = [1], pos = -1
Output: false
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: 165 ms)

Metric	Value
Runtime	165 ms
Memory	N/A
Date	2017-09-22

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 bool HasCycle(ListNode head) {
        if(head==null || head.next == null) return false;

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

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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

• Start by clarifying edge cases: empty input, single element, all duplicates.
• 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.