4.09.2023
141. Linked List Cycle easy
blog post
substack
Problem TLDR
Detect a cycle in a LinkedList
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 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
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, 104]
.-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?
Intuition
Use tortoise and rabbit technique
Approach
Move one pointer one step at a time, another two steps at a time. If there is a cycle, they will meet.
Complexity
Time complexity:
O(n)Space complexity:
O(log(n)) for recursion (iterative version is O(1))
Code
fun hasCycle(slow: ListNode?, fast: ListNode? = slow?.next): Boolean =
fast != null && (slow == fast || hasCycle(slow?.next, fast?.next?.next))