Question

In python, how do you move the pointer of a doubly linked list to the last node? Also in python, how do you move the current pointer "n" elements ahead (circularly)

def go_last(self):
# moves 'current' pointer to the last node

def skip(self,n:int):
# moves 'current' pointer n elements ahead (circularly)

Answer 1

Getting to the next node is what we have to do in both the part of the question,

The common part is the node and the doubly linked list class shown below:

class Node:                                     #this class objects are called for node creation
   def __init__(self, data):
      self.data = data
      self.next = None
      self.prev = None

class doubly_linked:                            #to create a doubly linked list
   def __init__(self):
      self.head = None

# not to get to the last node we have to do the same as tracing and store the previous node till null is the next node

 def go_last(self, node):
      while (node is not None):
         last = node
         node = node.next

# when we finish running the above code we'll have the last node stored in the variable last

# now in case of circular list there is no null so we have to change the looping condition for example,

n = int(input(" how many nodes does it has to jump"))

   def skip(self,node,n):
      while (n>0):
         last = node
         node = node.next
         n = n -1

# after n iterations the loop will end and the last variable will have the nth node

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Hope it helped