The Problem

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 the 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.

Examples

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).


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.

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.

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

Brute Force Solution

func hasCycle(_ head: ListNode?) -> Bool {
    var seen: Set<ListNode> = []
    var curr = head

    while curr != nil {
        if seen.contains(curr!) {
            return true
        }
        seen.insert(curr!)
        curr = curr?.next
    }

    return false
}

Explanation

We can solve this problem in a brute-force way using additional memory with a Set. By iterating through all nodes in head and checking if seen contains the current node, we can determine the result.

Time/Space Complexity

  • Time complexity: O(n)
  • Space complexity: O(n)

Slow/Fast Pointers Solution

func hasCycle(_ head: ListNode?) -> Bool {
    var slow = head
    var fast = head

    while fast != nil && fast?.next != nil {
        slow = slow?.next
        fast = fast?.next?.next
        if slow === fast {
            return true
        }
    }

    return false
}

Explanation

The slow/fast pointer technique solves the problem with constant memory space. We start from the beginning of head and move the slow pointer by one step and the fast pointer by two steps.
For example, in the case of input [1, 2, 3, 4] with pos = 1 (which corresponds to the value 2), the fast and slow pointers will eventually meet at the same position with value 2.

Time/Space Complexity

  • Time complexity: O(n)
  • Space complexity: O(1)

Thank you for reading! 😊