Find first node of loop in a linked list in python

Python program for Find first node of loop in a linked list. Here problem description and explanation.

#  Python 3 program for
#  Find first node of loop in a linked list

#  Node of Linked List
class LinkNode :
	def __init__(self, data) :
		#  Set node value
		self.data = data
		self.next = None
	

class SingleLL :
	def __init__(self) :
		self.head = None
		self.tail = None
		self.result = None
	
	#  Add new node at the end of linked list
	def insert(self, value) :
		#  Create a new node
		node = LinkNode(value)
		if (self.head == None) :
			self.head = node
		else :
			self.tail.next = node
		
		self.tail = node
	
	#  Display linked list element
	def display(self) :
		if (self.head == None) :
			return
		
		temp = self.head
		#  iterating linked list elements
		while (temp != None) :
			print(temp.data , end = " → ")
			#  Visit to next node
			temp = temp.next
		
		print("null")
	
	#  Find first loop node in given linked list
	def findLoopNode(self, node) :
		if (node == None or node.next == None) :
			#  When get node in loop
			self.result = node
			return node
		
		#  Get next node
		nextNode = node.next
		#  important to control recursion
		node.next = None
		node.next = self.findLoopNode(nextNode)
		return node
	
	#  This method are handles the request the finding loop node
	def firstLoopNode(self) :
		if (self.tail == None or self.tail.next == None) :
			print("Loop not exist")
		else :
			self.result = None
			self.findLoopNode(self.head)
			print("First Loop Node Is : ", self.result.data)
		
	

def main() :
	sll = SingleLL()
	#  Add linked list node
	sll.insert(1)
	sll.insert(2)
	sll.insert(3)
	sll.insert(4)
	sll.insert(5)
	sll.insert(6)
	sll.insert(7)
	print("Linked List")
	#  1 → 2 → 3 → 4 → 5 → 6 → 7 → null
	#  Assume that initial have no loop
	sll.display()
	#  Test Case 1
	#  When no loop
	sll.firstLoopNode()
	#  Test Case 2
	#  When loop exists
	#  Create a new loop
	sll.tail.next = sll.head.next.next
	sll.firstLoopNode()
	#  Create a new loop
	sll.tail.next = sll.head.next.next.next.next
	sll.firstLoopNode()
	#  remove loop
	sll.tail.next = None

if __name__ == "__main__": main()

Output

Linked List
1 → 2 → 3 → 4 → 5 → 6 → 7 → null
Loop not exist
First Loop Node Is :  3
First Loop Node Is :  5