Question

Need help on question No.5

MISU (Annoyed that this means you don't have much flexibility about how to implement this function? I apologizel But I don't regret it - I think that these debug printouts will help most students.) 5 accordion(old head) This function takes an old list, removes every other node, and then returns the updated list. Old nodes, which are removed from the list, may simply become garbage; you don't need to return them in any way. When removing nodes from the list, you must remove the old head, the node 2 after the old head, and the node 2 after that, out to the end of the list. For instance, if the input list is like this: 2 10 -> 20 -> -1 -> "asdf" -> 13 -> "qwerty" -> 1024 then the list that you return must look like this 20 -> "asdf -> "qwerty" If the oli list contained ay single mode or Wimpetum an empty Tist insert at (old head new.node, pos)

In python

Must use recursion

Must use recursion

Must use recursion

Thank you

LinkedList

Answer 1

Here is the code for your problem.

# Linked List class
class Node:  
    def __init__(self, data):  
        self.data = data  
        self.next = None
# recursive function to delete every other node
def accordion(old_head):
    # checks if head is none, then returns none
    if old_head == None:
        return None
    # we have to return this new_head which is the next element of the old_head
    new_head = old_head.next
    # this is a recursive call to the function
    if new_head != None and new_head.next != None:
        new_head.next = accordion(new_head.next)
    # return new_head
    return new_head
# utility functions insert and print just to demonstrate the "accordian" function
def insert_LL(head, data):
    node = Node(data)
    node.next = head
    return node
def print_LL(head):
    while head != None:
        print(head.data, end = " ")
        head = head.next
    print()
# main function for testing
if __name__ == '__main__':
    # making the LL
    head = None
    head = insert_LL(head, 1024)    
    head = insert_LL(head, "qwerty")    
    head = insert_LL(head, 13)    
    head = insert_LL(head, "asfd")    
    head = insert_LL(head, -1)    
    head = insert_LL(head, 20)    
    head = insert_LL(head, 10)    
    # printing old ll
    print("Old Linked List:", end = "\t")
    print_LL(head)
    # creating new LL and printing it
    new_head = accordion(head)
    print("New Linked List:", end = "\t")
    print_LL(new_head)

Here is the screenshot of the code if the indentation is not clear.

1 2 3 4 5 = 6 7 8 9 == 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 # Linked List class class Node: def __init__(self, data): self.data data self.next None # recursive function to delete every other node def accordion(old_head): # checks if head is none, then returns none if old_head None: return None # we have to return this new_head which is the next element of the old_head new_head old_head.next # this is a recursive call to the function if new_head != None and new_head.next != None: new_head.next = accordion(new_head.next) # return new_head return new_head # utility functions insert and print just to demonstrate the "accordian" function def insert_LL (head, data): node = Node(data) node. next head return node def print_LL(head): while head != None: print(head.data, end head head.next print() # main function for testing if _main__': # making the LL head None head insert_LL (head, 1024) head insert_LL(head, "qwerty") head insert_LL(head, 13) head insert_LL(head, "asfd") head insert_LL(head, -1) head insert_LL(head, 20) head insert_LL(head, 10) # printing old ll print("old Linked List:", end "\t") print_L4(head) # creating new LL and printing it new_head accordion(head) print("New Linked List:", end = "\t") print_LL(new_head) name

Here is the output of the code.

old Linked List: New Linked List: 10 20 -1 asfd 13 qwerty 1024 20 asfd qwerty

Hope this helps. Please rate the answer if you like it.
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