def union(llist_1, llist_2):
# edge case, if either of the lists is null, we return the other list
if llist_1 is None or llist_1.head is None:
return llist_2
if llist_2 is None or llist_2.head is None:
return llist_1
# the final linked list which we will return
unioned_list = LinkedList()
# in union, we need to make sure that elements in list 1 aren't repeated in list 2
# so we need to use a set to ensure no duplicates
unioned_set = set()
# we loop through the first linked list, and add its elements to the set
current_node = llist_1.head
while current_node is not None:
unioned_set.add(current_node.value)
current_node = current_node.next
# we loop through the second linked list, and add its elements to the set
current_node = llist_2.head
while current_node is not None:
unioned_set.add(current_node.value)
current_node = current_node.next
# now we create the final linked list and return it
for value in unioned_set:
unioned_list.append(value)
return unioned_list
def intersection(llist_1, llist_2):
# edge case, if any of the lists is null, we return null
if llist_1 is None or llist_2 is None:
return None
if llist_1.head is None or llist_2.head is None:
return None
# the final linked list which we will return
intersected_list = LinkedList()
# in intersection, we need to store items which appear in both lists
# so we use a set to store the elements in list 1
set_1 = set()
# and another set to store the intersected elements
intersected_set = set()
# we loop through the first linked list, and add its elements to set 1
current_node = llist_1.head
while current_node is not None:
set_1.add(current_node.value)
current_node = current_node.next
# we loop through the second linked list
current_node = llist_2.head
while current_node is not None:
# if the value of the current node exists in list 1, we add it to our final set
if current_node.value in set_1:
intersected_set.add(current_node.value)
current_node = current_node.next
# now we add the intersected elements to our final linked list and return it
for value in intersected_set:
intersected_list.append(value)
# if the intersection is an empty list, we return None
# otherwise we return the list
return intersected_list if intersected_list.head is not None else None
from Linked_List import LinkedList
from Union import union
from Intersection import intersection
# Test case 1, 2 linked lists with elements in union and intersection
print("\nChecking TEST_CASE_1: 2 linked lists with elements in union and intersection\n")
linked_list_1 = LinkedList()
linked_list_2 = LinkedList()
element_1 = [3,2,4,35,6,65,6,4,3,21]
element_2 = [6,32,4,9,6,1,11,21,1]
for i in element_1:
linked_list_1.append(i)
for i in element_2:
linked_list_2.append(i)
print (union(linked_list_1,linked_list_2))
print (intersection(linked_list_1,linked_list_2))
'''
Checking TEST_CASE_1: 2 linked lists with elements in union and intersection
32 -> 65 -> 2 -> 35 -> 4 -> 6 -> 1 -> 9 -> 11 -> 3 -> 21 ->
4 -> 21 -> 6 ->
'''
# Test case 2, 2 linked lists with elements in union but no intersection
print("\nChecking TEST_CASE_2: 2 linked lists with elements in union but no intersection\n")