LeetCode #142 — MEDIUM

Linked List Cycle II

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

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, 10⁴].
-10⁵ <= Node.val <= 10⁵
pos is -1 or a valid index in the linked-list.

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Linked List · Two Pointers

Example 1

[3,2,0,-4]
1

Example 2

[1,2]
0

Example 3

[1]
-1

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #142: Linked List Cycle II
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode fast = head, slow = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                ListNode ans = head;
                while (ans != slow) {
                    ans = ans.next;
                    slow = slow.next;
                }
                return ans;
            }
        }
        return null;
    }
}

// Accepted solution for LeetCode #142: Linked List Cycle II
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func detectCycle(head *ListNode) *ListNode {
	fast, slow := head, head
	for fast != nil && fast.Next != nil {
		slow = slow.Next
		fast = fast.Next.Next
		if slow == fast {
			ans := head
			for ans != slow {
				ans = ans.Next
				slow = slow.Next
			}
			return ans
		}
	}
	return nil
}

# Accepted solution for LeetCode #142: Linked List Cycle II
# 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

// Accepted solution for LeetCode #142: Linked List Cycle II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #142: Linked List Cycle II
// /**
//  * Definition for singly-linked list.
//  * class ListNode {
//  *     int val;
//  *     ListNode next;
//  *     ListNode(int x) {
//  *         val = x;
//  *         next = null;
//  *     }
//  * }
//  */
// public class Solution {
//     public ListNode detectCycle(ListNode head) {
//         ListNode fast = head, slow = head;
//         while (fast != null && fast.next != null) {
//             slow = slow.next;
//             fast = fast.next.next;
//             if (slow == fast) {
//                 ListNode ans = head;
//                 while (ans != slow) {
//                     ans = ans.next;
//                     slow = slow.next;
//                 }
//                 return ans;
//             }
//         }
//         return null;
//     }
// }

// Accepted solution for LeetCode #142: Linked List Cycle II
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function detectCycle(head: ListNode | null): ListNode | null {
    let [slow, fast] = [head, head];
    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
        if (slow === fast) {
            let ans = head;
            while (ans !== slow) {
                ans = ans.next;
                slow = slow.next;
            }
            return ans;
        }
    }
    return null;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.