142. Linked List Cycle II

https://leetcode.com/problems/linked-list-cycle-ii/

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: 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.

Follow-up:
Can you solve it without using extra space?

---

Intuition

Brute force / naive solution is to maintain a set of visited nodes.

If the set already contains current node, thats beginning of cycle => return that node

If pointer comes to end of list / null, there's no cycle

---

If there's a cycle, and two pointers are running at different speeds, they will meet at some point. This point proves that there's a cycle, and this point need not be start of cycle

Start another pointer at head once cycle is detected, they will meet at beginning of cycle
---
Time - O(n + k) - n = non cyclical length, k = cycle length
Space - O(1)
---
Related problems
linked-list-cycle
find-duplicate-number
---

---