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Linked List Cycle

Easy
Fast & Slow Pointers

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.

This is the classic application of Floyd's Tortoise and Hare algorithm. The slow pointer moves one step at a time, while the fast pointer moves two steps. If there's a cycle, the fast pointer will eventually catch up to the slow pointer.

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 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
Show Hints (4)
Hint 1: If two runners start at the same point and run around a circular track, will the faster runner ever lap the slower one?
Hint 2: Think about what happens when the fast pointer enters the cycle while the slow pointer is still outside.
Hint 3: Once both pointers are in the cycle, the distance between them decreases by 1 with each step. Why?
Hint 4: Can you prove that the fast pointer will always catch the slow pointer before the slow pointer completes one full cycle?
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