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.

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 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.

Source: leetcode | Difficulty: EASY
Tags: Data structures

Amortized analysis

Starter code: # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        

Solutions:
class Solution:

	def hasCycle(self, head: ListNode) -> bool:
		if head == None:
			return False
		slow = head
		fast = head.next
		while slow != fast:
			if fast is None or fast.next is None:
				return False
			slow = slow.next
			fast = fast.next.next
		return True


class Solution:

	def hasCycle(self, head: ListNode) -> bool:
		slow = head
		fast = head
		while fast and fast.next:
			slow = slow.next
			fast = fast.next.next
			if slow == fast:
				return True
		return False


class Solution:

	def hasCycle(self, head: ListNode) -> bool:
		nodes_seen = set()
		while head is not None:
			if head in nodes_seen:
				return True
			nodes_seen.add(head)
			head = head.next
		return False


class Solution:

	def hasCycle(self, head: ListNode) -> bool:
		cur = head
		check = set()
		while cur:
			if cur in check:
				return True
			check.add(cur)
			cur = cur.next
		return False


class Solution:

	def hasCycle(self, head: ListNode) -> bool:
		try:
			slow = head
			fast = head.next
			while slow is not fast:
				slow = slow.next
				fast = fast.next.next
			return True
		except:
			return False