Nth node from end of linked list Python

PROGRAM TO GET Nth NODE FROM END OF LINKED LIST

# Node class 

class Node:

  

    # Constructor to initialize the node object

    def __init__(self, data):

        self.data = data

        self.next = None

  

class LinkedList:

  

    # Function to initialize head

    def __init__(self):

        self.head = None

  

    # Function to insert a new node at the beginning

    def push(self, new_data):

        new_node = Node(new_data)

        new_node.next = self.head

        self.head = new_node

  

    def printNthFromLast(self, n):

        main_ptr = self.head

        ref_ptr = self.head 

      

        count = 0 

        if(self.head is not None):

            while(count < n ):

                if(ref_ptr is None):

                    print "% d is greater than the 

                           no. pf nodes in list" %(n)

                    return

   

                ref_ptr = ref_ptr.next

                count += 1

      

        if(ref_ptr is None):

            self.head = self.head.next

            if(self.head is not None):

                 print "Node no. % d from last is % d " 

                                   %(n, main_ptr.data)

        else:

            

  

          while(ref_ptr is not None):

              main_ptr = main_ptr.next 

              ref_ptr = ref_ptr.next

  

          print "Node no. % d from last is % d " 

                                     %(n, main_ptr.data)

  

  

# Driver program to test above function

llist = LinkedList()

llist.push(20)

llist.push(4)

llist.push(15)

llist.push(35)

  

llist.printNthFromLast(4)




OUTPUT

Node no. 4 from last is 35 

CREDITS: https://www.geeksforgeeks.org/nth-node-from-the-end-of-a-linked-list/