LinkedListNthToLastNode.py

#  Linked list Nth to last node

# Author: Pradeep K. Pant, ppant@cpan.org
# Aim is a function that takes a head node and an integer value n and then 
# returns the nth to last node in the linked list.

# First initialize singly linked list class
class LinkedListNode(object):
    def __init__(self, value):
        self.value = value
        self.nextnode = None
        
def nth_to_last_node(n, head):
    # Set variables
    # Left pointer at head and right pointer at place nth distance from
    # left pointer
    # so we have blocks between left and right pointers we keep them moving
    # till right pointer hits the tail
    left_pointer = head
    right_pointer = head

    # setting up right pointer based on the val of n
    for i in range(n - 1):
        # Check edge cases
        if not right_pointer.nextnode:
            raise LookupError('Error: n is larger than linked list')
        right_pointer = right_pointer.nextnode

    # finally check if right pointer is hitting the tail if so return left pointer
    while right_pointer.nextnode:
        left_pointer = left_pointer.nextnode
        right_pointer = right_pointer.nextnode
    return left_pointer

# Test
# Create a Linked List 
a = LinkedListNode(1)
b = LinkedListNode(2)
c = LinkedListNode(3)
d = LinkedListNode(4)
e = LinkedListNode(5)

a.nextnode = b
b.nextnode = c
c.nextnode = d
d.nextnode = e

print (f"a.nextnode: {a.nextnode.value}")
print (f"b.nextnode: {b.nextnode.value}")
print (f"c.nextnode: {c.nextnode.value}")
print (f"d.nextnode: {d.nextnode.value}")

# This would return the node d with a value of 4, because its the 2nd to last node.
target_node = nth_to_last_node(2, a)
print (f"Target node value (2nd to last): {target_node.value}")
# Ans: d=4