cc 2.2

nthtolast.py

#2.2 Implement an algorithm to find the kth to last element of a singly linked list.
# supose that the length of the link is unknown

from LinkedList import Node, LinkedList

def nthtolast(head, k):
    
    if head is None:
        return  
    p1 = head
    p2 = head
    for i in range(0, k-1):
        if p2 is not None:
            p2 =p2.next
    while p2.next is not None:
        p1 = p1.next
        p2 = p2.next
    print p1.data
  
                     
test = LinkedList()
for i in range(15):
    test.AddNode(i)

nthtolast(test.head, 5)