class LinkedListTests(unittest.TestCase):
def setUp(self):
self.list = LinkedList()
def test_add_to_tail(self):
self.list.add_to_tail(1)
self.assertEqual(self.list.tail.value, 1)
self.assertEqual(self.list.head.value, 1)
self.list.add_to_tail(2)
self.assertEqual(self.list.tail.value, 2)
self.assertEqual(self.list.head.value, 1)
def test_contains(self):
self.list.add_to_tail(1)
self.list.add_to_tail(2)
self.list.add_to_tail(5)
self.list.add_to_tail(10)
self.assertTrue(self.list.contains(10))
self.assertTrue(self.list.contains(2))
self.assertFalse(self.list.contains(1000))
def test_remove_head(self):
self.list.add_to_tail(10)
self.list.add_to_tail(20)
self.assertEqual(self.list.remove_head(), 10)
self.assertFalse(self.list.contains(10))
self.assertEqual(self.list.remove_head(), 20)
self.assertFalse(self.list.contains(20))
self.list.add_to_tail(10)
self.assertEqual(self.list.remove_head(), 10)
self.assertIsNone(self.list.head)
self.assertIsNone(self.list.tail)
self.assertIsNone(self.list.remove_head())
# if name_1 == name_2:
# duplicates.append(name_1)
#########
# Using comprehensions
# duplicates = [name for name in names_1 if name in names_2]
########
# Using Linked List
# create a new list
names_1_list = LinkedList()
# put names_1 in list
for name in names_1:
names_1_list.add_to_head(name)
# Check to see if each item of list 2 is in list 1
for name in names_2:
if names_1_list.contains(name):
duplicates.append(name)
end_time = time.time()
print(f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
print(f"runtime: {end_time - start_time} seconds")
# ---------- Stretch Goal -----------
# Python has built-in tools that allow for a very efficient approach to this problem
# What's the best time you can accomplish? There are no restrictions on techniques or data
# structures, but you may not import any additional libraries that you did not write yourself.
