Python Program For Checking Linked List With A Loop Is Palindrome Or Not

Last Updated : 19 Jul, 2022

Given a linked list with a loop, the task is to find whether it is palindrome or not. You are not allowed to remove the loop.

Examples:

Input: 1 -> 2 -> 3 -> 2              /|      |/                ------- 1   Output: Palindrome Linked list is 1 2 3 2 1 which is a  palindrome.  Input: 1 -> 2 -> 3 -> 4              /|      |/                ------- 1   Output: Not Palindrome Linked list is 1 2 3 4 1 which is a  not palindrome.

Algorithm:

Detect the loop using the Floyd Cycle Detection Algorithm.
Then find the starting node of the loop as discussed in this.
Check the linked list is palindrome or not as discussed in this.

Below is the implementation.

Python

# Python3 program to check if a  # linked list with loop is palindrome  # or not.  # Node class  class Node:       # Constructor to initialize the      # node object      def __init__(self, data):          self.data = data          self.next = None  # Function to find loop starting node.  # loop_node -. Pointer to one of  # the loop nodes head -. Pointer to  # the start node of the linked list def getLoopstart(loop_node,head):       ptr1 = loop_node      ptr2 = loop_node       # Count the number of nodes in      # loop      k = 1     i = 0     while (ptr1.next != ptr2):              ptr1 = ptr1.next         k = k + 1          # Fix one pointer to head      ptr1 = head       # And the other pointer to k      # nodes after head      ptr2 = head     i = 0     while (i < k):         ptr2 = ptr2.next         i = i + 1      # Move both pointers at the same pace,      # they will meet at loop starting node      while (ptr2 != ptr1):              ptr1 = ptr1.next         ptr2 = ptr2.next          return ptr1   # This function detects and find  # loop starting node in the list def detectAndgetLoopstarting(head):      slow_p = head     fast_p = head     loop_start = None      # Start traversing list and detect loop      while (slow_p != None and fast_p != None and            fast_p.next != None):             slow_p = slow_p.next         fast_p = fast_p.next.next          # If slow_p and fast_p meet then find          # the loop starting node         if (slow_p == fast_p):                     loop_start = getLoopstart(slow_p,                                        head)              break              # Return starting node of loop      return loop_start   # Utility function to check if  # a linked list with loop is  # palindrome with given starting point.  def isPalindromeUtil(head, loop_start):      ptr = head      s = []       # Traverse linked list until last node      # is equal to loop_start and store the      # elements till start in a stack      count = 0     while (ptr != loop_start or count != 1):              s.append(ptr.data)          if (ptr == loop_start) :             count = 1         ptr = ptr.next          ptr = head      count = 0      # Traverse linked list until last node is      # equal to loop_start second time      while (ptr != loop_start or count != 1):              # Compare data of node with the top of stack          # If equal then continue          if (ptr.data == s[-1]):              s.pop()           # Else return False          else:             return False          if (ptr == loop_start) :             count = 1         ptr = ptr.next          # Return True if linked list is      # palindrome      return True  # Function to find if linked list # is palindrome or not  def isPalindrome(head):      # Find the loop starting node      loop_start =      detectAndgetLoopstarting(head)       # Check if linked list is palindrome      return isPalindromeUtil(head,                              loop_start)   def newNode(key):     temp = Node(0)      temp.data = key      temp.next = None     return temp   # Driver code head = newNode(50)  head.next = newNode(20)  head.next.next = newNode(15)  head.next.next.next = newNode(20)  head.next.next.next.next = newNode(50)   # Create a loop for testing  head.next.next.next.next.next =  head.next.next  if(isPalindrome(head) == True):     print("Palindrome") else:     print("Not Palindrome")  # This code is contributed by Arnab Kundu

Output:

Palindrome

Time Complexity: O(n) where n is no of nodes in the linked list

Auxiliary Space: O(n)

Please refer complete article on Check linked list with a loop is palindrome or not for more details!

Python Program For Finding The Length Of Loop In Linked List

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Python Program For Checking Linked List With A Loop Is Palindrome Or Not

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