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Lilly C16 Pine #64

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163 changes: 146 additions & 17 deletions linked_list/linked_list.py
Original file line number Diff line number Diff line change
Expand Up @@ -13,63 +13,140 @@ def __init__(self):

# returns the value in the first node
# returns None if the list is empty
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(1)?
# Space Complexity: O(1)?
def get_first(self):

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👍

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resubmitted with questions filled out

pass
if self.head == None:
return None

return self.head.value

# method to add a new node with the specific data value in the linked list
# insert the new node at the beginning of the linked list
# Time Complexity: ?
# Space Complexity: ?
def add_first(self, value):

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👍

pass
self.head = Node(value, self.head)

# method to find if the linked list contains a node with specified value
# returns true if found, false otherwise
# Time Complexity: ?
# Space Complexity: ?
def search(self, value):
pass
current = self.head

while current != None:

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We can shorten this to:

Suggested change
while current != None:
while current:

if current.value == value:
return True
current = current.next
return False

# method that returns the length of the singly linked list
# Time Complexity: ?
# Space Complexity: ?

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What do you think the space and time complexity of this is? Is the time complexity dependent on the length of the linked list? Is there any new variables being created that grow larger depending upon the input?

def length(self):

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👍

pass
current = self.head

count = 0
while current != None:

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Suggested change
while current != None:
while current:

current = current.next
count += 1

return count

# method that returns the value at a given index in the linked list
# index count starts at 0
# returns None if there are fewer nodes in the linked list than the index value
# Time Complexity: ?
# Space Complexity: ?

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What do you think the space and time complexity of this is? Is the time complexity dependent on the length of the linked list? Is there any new variables being created that grow larger depending upon the input?

def get_at_index(self, index):

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👍

pass
current = self.head

count = 0
while current:
if count == index:
return current.value
count += 1
current = current.next

return None

# method that returns the index of the first node with the given value
# returns -1 if the value is not found in the linked list
# Time Complexity: O(n)?
# Space Complexity: O(1)?


# method that returns the value of the last node in the linked list
# returns None if the linked list is empty
# Time Complexity: ?
# Space Complexity: ?
def get_last(self):

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👍

pass

if self.head == None:
return None

current = self.head

while current.next != None:
current = current.next

return current.value

# method that inserts a given value as a new last node in the linked list
# Time Complexity: ?
# Space Complexity: ?

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What do you think the space and time complexity of this is? Is the time complexity dependent on the length of the linked list? Is there any new variables being created that grow larger depending upon the input?

#method to add node to the end of the linked list
def add_last(self, value):

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👍

pass
current = self.head
if current is None:
self.head = Node(value)
return
while current.next != None:

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Suggested change
while current.next != None:
while current.next:

current = current.next
current.next = Node(value)


# method to return the max value in the linked list
# returns the data value and not the node
def find_max(self):

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👍

pass
if self.head == None:
return None

current = self.head

max = 0
while current:
if current.value > max:
max = current.value
current = current.next

return max

# method to delete the first node found with specified value
# Time Complexity: ?
# Space Complexity: ?
Comment on lines 129 to 130

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What do you think the space and time complexity of this is? Is the time complexity dependent on the length of the linked list? Is there any new variables being created that grow larger depending upon the input?

def delete(self, value):
pass
current = self.head

if current is not None:
if current.value == value:
self.head = current.next
temp = None
return

Comment on lines +132 to +139

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We can shorten this up a bit to make it more readable:

Suggested change
current = self.head
if current is not None:
if current.value == value:
self.head = current.next
temp = None
return
if self.head.value == value:
self.head = current.next
return
current = self.head

while current is not None:
if current.value == value:
break
prev = current
current = current.next

if current == None:
return None

prev.next = current.next

current = None
Comment on lines +140 to +151

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We can shorten this, too, to make it more readable:

Suggested change
while current is not None:
if current.value == value:
break
prev = current
current = current.next
if current == None:
return None
prev.next = current.next
current = None
while current is not None:
if current.next.value == value:
current.next = current.next.next
current = current.next

We don't need to re-assign current to None nor return None


# method to print all the values in the linked list
# Time Complexity: ?
Expand All @@ -86,32 +163,84 @@ def visit(self):

# method to reverse the singly linked list
# note: the nodes should be moved and not just the values in the nodes
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)?
# Space Complexity: O(1)?
def reverse(self):

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👍

pass
prev = None
current = self.head
while current is not None:
next = current.next
current.next = prev
prev = current
current = next
self.head = prev

## Advanced/ Exercises
# returns the value at the middle element in the singly linked list
# Time Complexity: ?
# Space Complexity: ?
# find middle value of a singly linked list
def find_middle_value(self):
pass
if self.head == None:
return None

current = self.head
count = 0
while current:
count += 1
current = current.next
Comment on lines +189 to +193

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hmm where have we seen this already? Maybe we can call length to get the number of nodes


current = self.head
for i in range(count // 2):
current = current.next

return current.value



# find the nth node from the end and return its value
# assume indexing starts at 0 while counting to n
# Time Complexity: ?
# Space Complexity: ?
# find nth node from the end of the linked list
def find_nth_from_end(self, n):
pass
if self.head == None:
return None

current = self.head
count = 0
while current:
count += 1
current = current.next

current = self.head
for i in range(count - n):
current = current.next

return current.value




# checks if the linked list has a cycle. A cycle exists if any node in the
# linked list links to a node already visited.
# returns true if a cycle is found, false otherwise.
# Time Complexity: ?
# Space Complexity: ?

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What do you think the space and time complexity of this is? Is the time complexity dependent on the length of the linked list? Is there any new variables being created that grow larger depending upon the input?

def has_cycle(self):

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👍 interesting approach!

pass
if self.head == None:
return False

current = self.head
visited = set()

while current:
if current in visited:
return True
visited.add(current)
current = current.next

return False

# Helper method for tests
# Creates a cycle in the linked list for testing purposes
Expand Down
12 changes: 6 additions & 6 deletions tests/linked_list_test.py
Original file line number Diff line number Diff line change
Expand Up @@ -197,7 +197,7 @@ def test_reverse_will_reverse_five_element_list(list):
for i in range(0, 5):
assert list.get_at_index(i) == i

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_find_middle_value_returns_middle_element_of_five_element_list(list):
list.add_first(10)
list.add_first(30)
Expand All @@ -206,7 +206,7 @@ def test_find_middle_value_returns_middle_element_of_five_element_list(list):
list.add_first(20)
assert list.find_middle_value() == 50

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_find_middle_value_returns_element_at_index_two_of_six_element_list(list):
list.add_first(10)
list.add_first(30)
Expand All @@ -216,11 +216,11 @@ def test_find_middle_value_returns_element_at_index_two_of_six_element_list(list
list.add_first(100)
assert list.find_middle_value() == 60

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_nth_from_n_when_list_is_empty(list):
assert list.find_nth_from_end(3) == None

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_find_nth_from_n_when_length_less_than_n(list):
list.add_first(5)
list.add_first(4)
Expand All @@ -230,7 +230,7 @@ def test_find_nth_from_n_when_length_less_than_n(list):

assert list.find_nth_from_end(6) == None

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_find_nth_from_n(list):
list.add_first(1)
list.add_first(2)
Expand All @@ -243,7 +243,7 @@ def test_find_nth_from_n(list):
assert list.find_nth_from_end(3) == 4
assert list.find_nth_from_end(4) == None

@pytest.mark.skip(reason="Going Further methods")
# @pytest.mark.skip(reason="Going Further methods")
def test_has_cycle(list):
assert list.has_cycle() == False

Expand Down