def test_more_complicated(self):
start_node = NodeList(1, NodeList(2, NodeList(3, NodeList(4, NodeList(1, None)))))
self.assertEqual(5, node_len(start_node))
actual_list = node_remove_dups(start_node)
self.assertEqual(4, node_len(actual_list))
def test_one_node(self):
start_node = NodeList(1, None)
actual_list = node_remove_dups(start_node)
self.assertEqual(1, node_len(actual_list))
def test_double_2_duplicate(self):
start_node = NodeList(1, NodeList(1, NodeList(1, None)))
actual_list = node_remove_dups(start_node)
self.assertEqual(1, node_len(actual_list))
def sum_lists_big_to_small(first_list_start, second_list_start):
first_len = node_len(first_list_start)
second_len = node_len(second_list_start)
# this will be used to shift the smaller one
difference_len = abs(first_len - second_len)
if first_len > second_len:
second_list_start = pad_with_zero(second_list_start, difference_len)
else:
first_list_start = pad_with_zero(first_list_start, difference_len)
result_list = recurrence_sum(first_list_start, second_list_start)
# due to recurence sum imperfection if first elem is 0 then discard it and shift to next
if result_list.val == 0:
result_list = result_list.next
return result_list
def brute_force(start_node, kth):
length = node_len(start_node) # O(n)
if length - kth <= 0:
return None
counter = 0
current_node = start_node
while current_node != None:
counter += 1
if counter >= length - kth:
return current_node
current_node = current_node.next
return None
def is_palindrome(list_start):
l = node_len(list_start)
_, result = is_palindrome_recursive(list_start, l)
return result