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Mary Tian CS Fun Linked Lists #80

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157 changes: 116 additions & 41 deletions linked_list/linked_list.py
Original file line number Diff line number Diff line change
Expand Up @@ -7,73 +7,146 @@ def __init__(self, value, next_node = None):
self.next = next_node

# Defines the singly linked list


class LinkedList:
def __init__(self):
self.head = None # keep the head private. Not accessible outside this class

# 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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pass
if self.head is 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: ?
# Time Complexity: O(1)
# Space Complexity: O(1)
def add_first(self, value):

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pass
new_node = Node(value)
new_node.next = self.head
self.head = new_node

return self.head.value

# method to find if the linked list contains a node with specified value
# returns true if found, false otherwise
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def search(self, value):

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pass
current = self.head

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: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def length(self):

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pass
current = self.head
count = 0

while current != None:
count += 1
current = current.next

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: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def get_at_index(self, index):

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pass
i = 0
current = self.head

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

return None


# 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: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def get_last(self):

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pass
current = self.head

if not self.head:
return None

# keep looping until end of the links
while current.next:
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: ?
# Time Complexity: O(1)

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⏱ Time complexity here would be O(n) because we always have to traverse the entire list to get to the tail. With a doubly linked list where you maintain a tail pointer, this can become an O(1) operation.

# Space Complexity: O(1)
def add_last(self, value):
pass
current = self.head

if current is None:
self.head = Node(value, None)
else:
while current:
if current.next == None:
break
Comment on lines +107 to +108

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🤓 We generally like to avoid break statements where possible. How might you refactor your code to eliminate this break statement?

current = current.next

current.next = Node(value, None)

# 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
current = self.head
if current is None:
return None
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: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def delete(self, value):

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pass
current = self.head
prev = None
if not current:
return None

while current:
if current.value == value:
if current == self.head:
self.head = current.next
return
prev.next = current.next
return
prev = current
current = current.next

# method to print all the values in the linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(n)
def visit(self):

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helper_list = []
current = self.head
Expand All @@ -86,24 +159,34 @@ 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
current = self.head
previous = None

while current:
next = current.next
current.next = previous
previous = current
current = next

self.head = previous

## Advanced/ Exercises
# returns the value at the middle element in the singly linked list
# Time Complexity: ?
# Space Complexity: ?
def find_middle_value(self):
pass
pass

# find the nth node from the end and return its value
# assume indexing starts at 0 while counting to n
# Time Complexity: ?
# Space Complexity: ?
def find_nth_from_end(self, n):
pass
pass


# checks if the linked list has a cycle. A cycle exists if any node in the
# linked list links to a node already visited.
Expand All @@ -117,12 +200,4 @@ def has_cycle(self):
# Creates a cycle in the linked list for testing purposes
# Assumes the linked list has at least one node
def create_cycle(self):
if self.head == None:
return

# navigate to last node
current = self.head
while current.next != None:
current = current.next

current.next = self.head # make the last node link to first node
pass