Question

PYTHON 3 LANGUAGE , LINKED LISTS , define function ITERATIVELY

Define an iterative function named alternate_i; it is passed two linked lists (ll1 and ll2) as arguments. It returns a reference to the front of a linked list that alternates the LNs from ll1 and ll2, starting with ll1; if either linked list becomes empty, the rest of the LNs come from the other linked list. So, all LNs in ll1 and ll2 appear in the returned result (in the same relative order, possibly separated by values from the other linked list). The original linked lists are mutated by this function (the .nexts are changed; create no new LN objects). For example, if we defined

a = list_to_ll(['a', 'b', 'c',]) and b = list_to_ll([1,2,3,4,5])

alternate_i(a,b) returns a->1->b->2->c->3->4->5->None and alternate_i(b,a) returns 1->a->2->b->3- >c->4->5->None. You may not create/use any Python data structures in your code: use linked list processing only. Change only .next attributes (not .value attributes).

def alternate_i(ll1 : LN, ll2 : LN) -> LN:
# Handle the case of ll1 or ll2 being empty
# Set up for iteration (keep track of front and rear of linked list to retur
# while True: with both l1 and l2 not empty and ll1 is in the linked list to return
while True:
break
# return correct result if ll1 or ll2 is empty (after advancing ll1 and ll2)
# continue looping if both ll1/ll2 are not empty, with ll1 in the linked list to return

Answer 1

def alternate_i(ll1, ll2):
  
   # if first list is empty
   if ll1 is None:
       return ll2
  
   # if second list is empty
   elif ll2 is None:
       return ll1
  
   # if both list are empty
   else:
       return None
      
   prev = None # temp variable
  
   # loop untill both list are empty
   while ll1 not None and ll2 not None:
       prev = ll1.next # store next address of first list
       ll1.next = ll2 # add the current element of second list to first
       ll2 = ll2.next # increase the second list
       ll1.next.next = prev # add the remaining element of first list back
       ll1 = ll1.next.next # increment first list
  
   # if first list contain less no of elements than second add full second list
   if ll2 not None:
       prev.next = ll2
  
   return ll1