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Leetcode #142: Linked List Cycle II

In this guide, we solve Leetcode #142 Linked List Cycle II in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Hash Table, Linked List, Two Pointers

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

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.

Python Solution

# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]: fast = slow = head while fast and fast.next: slow = slow.next fast = fast.next.next if slow == fast: ans = head while ans != slow: ans = ans.next slow = slow.next return ans

Complexity

The time complexity is O(n)O(n)O(n), where nnn is the number of nodes in the linked list. The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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