Python Program For Finding The Length Of Loop In Linked List

Write a function detectAndCountLoop() that checks whether a given Linked List contains loop and if loop is present then returns count of nodes in loop. For example, the loop is present in below-linked list and length of the loop is 4. If the loop is not present, then the function should return 0.

Approach:
It is known that Floyd’s Cycle detection algorithm terminates when fast and slow pointers meet at a common point. It is also known that this common point is one of the loop nodes. Store the address of this common point in a pointer variable say (ptr). Then initialize a counter with 1 and start from the common point and keeps on visiting the next node and increasing the counter till the common pointer is reached again. 
At that point, the value of the counter will be equal to the length of the loop.
Algorithm:

1. Find the common point in the loop by using the Floyd’s Cycle detection algorithm
2. Store the pointer in a temporary variable and keep a count = 0
3. Traverse the linked list until the same node is reached again and increase the count while moving to next node.
4. Print the count as length of loop

Python3

# Python program to detect a loop and
# find the length of the loop
# Node defining class
class Node:

    # Function to make a node
    def __init__(self, val):
        self.val = val
        self.next = None

# Linked List defining and loop
# length finding class
class LinkedList:

    # Function to initialize the
    # head of the linked list
    def __init__(self):
        self.head = None

    # Function to insert a new
    # node at the end
    def AddNode(self, val):
        if self.head is None:
            self.head = Node(val)
        else:
            curr = self.head
            while(curr.next):
                curr = curr.next
            curr.next = Node(val)

    # Function to create a loop in the
    # Linked List. This function creates
    # a loop by connecting the last node
    # to n^th node of the linked list,
    # (counting first node as 1)
    def CreateLoop(self, n):

        # LoopNode is the connecting node to
        # the last node of linked list
        LoopNode = self.head
        for _ in range(1, n):
            LoopNode = LoopNode.next

        # end is the last node of the
        # Linked List
        end = self.head
        while(end.next):
            end = end.next

        # Creating the loop
        end.next = LoopNode

    # Function to detect the loop and return
    # the length of the loop if the returned
    # value is zero, which means that either
    # the linked list is empty or the linked
    # list doesn't have any loop
    def detectLoop(self):

        # If linked list is empty then there
        # is no loop, so return 0
        if self.head is None:
            return 0

        # Using Floyd’s Cycle-Finding
        # Algorithm/ Slow-Fast Pointer Method
        slow = self.head
        fast = self.head

        # To show that both slow and fast
        # are at start of the Linked List
        flag = 0

        while(slow and slow.next and fast and
              fast.next and fast.next.next):
            if slow == fast and flag != 0:

                # Means loop is confirmed in the
                # Linked List. Now slow and fast
                # are both at the same node which
                # is part of the loop
                count = 1
                slow = slow.next
                while(slow != fast):
                    slow = slow.next
                    count += 1
                return count

            slow = slow.next
            fast = fast.next.next
            flag = 1
        return 0 # No loop

# Setting up the code
# Making a Linked List and
# adding the nodes
myLL = LinkedList()
myLL.AddNode(1)
myLL.AddNode(2)
myLL.AddNode(3)
myLL.AddNode(4)
myLL.AddNode(5)

# Creating a loop in the linked List
# Loop is created by connecting the
# last node of linked list to n^th node
# 1<= n <= len(LinkedList)
myLL.CreateLoop(2)

# Checking for Loop in the Linked List
# and printing the length of the loop
loopLength = myLL.detectLoop()
if myLL.head is None:
    print("Linked list is empty")
else:
    print(str(loopLength))
# This code is contributed by _Ashutosh

Output:

4

Complexity Analysis:

• Time complexity:O(n). 
  Only one traversal of the linked list is needed.
• Auxiliary Space:O(1). 
  As no extra space is required.

Please refer complete article on Find length of loop in linked list for more details!