def bst_sequences(root):
"""
This is a Python implementation of Gayle's solution from CtCI.
I struggled with this problem so I just want to move past it.
"""
result = LinkedList()
if root is None:
result.insert(LinkedList())
return result
prefix = LinkedList()
prefix.add(root.data)
#Recurse on left and right subtrees
left_sequences = bst_sequences(root.left)
right_sequences = bst_sequences(root.right)
#Weave together each list from the left and right sides.
while not left_sequences.is_empty():
left = left_sequences.get_first().data
while not right_sequences.is_empty():
right = right_sequences.get_first().data
weaved = LinkedList()
weave_lists(left, right, weaved, prefix)
result.add_all(weaved)
return result
def sum_lists(a, b):
a_num = 0
a_mult = 1
a_curr = a.head
while a_curr is not None:
a_num += a_curr.data * a_mult
a_curr = a_curr._next
a_mult *= 10
b_num = 0
b_mult = 1
b_curr = b.head
while b_curr is not None:
b_num += b_curr.data * b_mult
b_curr = b_curr._next
b_mult *= 10
output_num = a_num + b_num
print(output_num)
output_mult = 1
while output_num // (output_mult * 10) > 0:
output_mult *= 10
output = LinkedList()
if output_num == 0:
output.insert(0)
return output
while output_num != 0:
digit = int(output_num // output_mult)
output.insert(digit)
output_num = output_num - (digit * output_mult)
output_mult /= 10
return output
def test_sum_lists(num_a, num_b):
a = LinkedList()
a_str = str(num_a)
for digit in a_str:
a.insert(int(digit))
b = LinkedList()
b_str = str(num_b)
for digit in b_str:
b.insert(int(digit))
a.print_list()
b.print_list()
result = sum_lists(a, b)
result.print_list()
def is_a_palindrome(li):
"""
This solution relies on the fact that the linked list
implementation inserts nodes at the head of the list.
It first scans the list to count the number of elements.
Then we pop off the first half of the list while inserting
into a new list. This will reverse the elements in a stack-like
manner. If the original list contains an odd number
of elements we discard the middle element. Finally, we
check to see if the two lists we're left with match.
This solution destroyes the original list, we could
easily add a step to copy it into a new list first to avoid
destroying it.
"""
n = 0
curr = li.head
while curr is not None:
n += 1
curr = curr._next
li2 = LinkedList()
k = n // 2
while k > 0:
popped_node = li.get_first()
li2.insert(popped_node.data)
k -= 1
if n % 2 == 1:
#Remove the middle element and discard it.
_ = li.get_first()
while not li2.is_empty():
li2_el = li2.get_first()
li_el = li.get_first()
if li2_el.data != li_el.data:
return False
return True
def test_loop_detection():
li = LinkedList()
li.insert('C')
li.insert('E')
li.insert('D')
c_node = li.head._next._next
c_node._next = li.head
li.head = c_node
#Oops now we have a loop.
li.insert('B')
li.insert('A')
loop_node = loop_detection(li)
if loop_node is not None:
print('Has loop at node %s.' % loop_node.data)
else:
print('Contains no loop.')
class TestLinkedList(unittest.TestCase):
linked_list = None
def setUp(self):
self.linked_list = LinkedList()
def test_init(self):
self.assertEqual(self.linked_list.head, None,
"Initial HEAD should be None")
self.assertEqual(len(self.linked_list), 0,
"Initial length should be zero")
def test_len(self):
self.assertEqual(len(self.linked_list), 0,
"Newly initiated list's length should be zero")
self.linked_list.insert(1)
self.assertEqual(len(self.linked_list), 1, "List length should be 1")
def test_list(self):
node1 = self.linked_list.Node(1, None)
node2 = self.linked_list.Node(2, node1)
node3 = self.linked_list.Node(3, node2)
self.linked_list.insert(1)
self.linked_list.insert(2)
self.linked_list.insert(3)
self.assertEqual([node3, node2, node1], list(self.linked_list))
def test_str(self):
self.linked_list.insert(1)
self.linked_list.insert(2)
self.linked_list.insert(3)
self.assertEqual(str(self.linked_list), '[3, 2, 1]',
"List should be printed as [3, 2, 1]")
self.linked_list.delete(2)
self.assertEqual(str(self.linked_list), '[3, 1]',
"List should be printed as [3, 1]")
self.linked_list.delete(3)
self.assertEqual(str(self.linked_list), '[1]',
"List should be printed as [1]")
self.linked_list.delete(1)
self.assertEqual(str(self.linked_list), '[]',
"List should be printed as []")
def test_insert(self):
self.assertEqual(
self.linked_list.insert(1), self.linked_list.Node(1, None),
"Inserting 1 into list should return node with value=1")
self.assertEqual(list(self.linked_list), [self.linked_list.Node(1)],
"Inserting 1 into empty list should give [1]")
self.linked_list.insert(3, 1)
self.assertEqual(self.linked_list.head.next,
self.linked_list.Node(3, None),
"Inserting 3 into pos=1 of [1] should give [1,3]")
self.linked_list.insert(2, 1)
self.assertEqual(
self.linked_list.head.next.value,
self.linked_list.Node(2, None).value,
"Inserting 2 into pos=1 of [1,3] should give [1,2,3]")
def test_contains(self):
self.linked_list.insert(1)
self.linked_list.insert(2)
self.linked_list.insert(3)
self.assertEqual(
1 in self.linked_list, True,
"After inserting 1 into the list, we should be able to find it there"
)
self.assertEqual(
4 in self.linked_list, False,
"After inserting 1 into the list, we should be able to find it there"
)
#print(self.linked_list)
def test_search(self):
self.linked_list.insert(1)
self.linked_list.insert(2)
self.linked_list.insert(3)
self.assertEqual(
self.linked_list.search(2).value,
self.linked_list.Node(2, None).value,
"Searching for 2 in [3,2,1] should return node with value=2")
self.assertEqual(self.linked_list.search(4), None,
"Searching for 4 in [3,2,1] should return None")
def test_delete(self):
self.linked_list.insert(1)
self.linked_list.insert(2)
self.linked_list.insert(3)
self.assertEqual(
self.linked_list.delete(2).value,
self.linked_list.Node(2, None).value,
"Deleting 2 from [3,2,1] should return the node with value 2")
self.linked_list.delete(3)
self.assertEqual(
self.linked_list.head, self.linked_list.Node(1, None),
"Deleting 2 and 3 from [3,2,1] should leave the list as [1]")
