Linked List Cycle

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.

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 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?

Solution

To solve this problem, allocate a slow and a fast pointer, where the slow pointer iterates through the list twice as fast as the slower pointer.

If they intersect before the slow pointer gets to the end, then there is a cycle. Otherwise, there is no cycle.

This takes O(n) time (as long as it does to traverse the list) and O(1) space (two pointers).

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        if not head:
            return False
        slow = head
        fast = head.next

        while slow and fast and fast.next:
            if slow == fast:
                return True
            slow = slow.next
            fast = fast.next.next

        return False

Table of Contents

Linked List Cycle

Solution