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Mary Tian CS Fun Linked Lists #80
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@@ -7,73 +7,146 @@ def __init__(self, value, next_node = None): | |
self.next = next_node | ||
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# Defines the singly linked list | ||
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class LinkedList: | ||
def __init__(self): | ||
self.head = None # keep the head private. Not accessible outside this class | ||
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# 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): | ||
pass | ||
if self.head is None: | ||
return None | ||
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return self.head.value | ||
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# 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): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
new_node = Node(value) | ||
new_node.next = self.head | ||
self.head = new_node | ||
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return self.head.value | ||
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# 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): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
current = self.head | ||
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while current: | ||
if current.value == value: | ||
return True | ||
current = current.next | ||
return False | ||
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# 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 | ||
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while current != None: | ||
count += 1 | ||
current = current.next | ||
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return count | ||
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# 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 | ||
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while current: | ||
if i == index: | ||
return current.value | ||
current = current.next | ||
i += 1 | ||
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return None | ||
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# 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 | ||
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if not self.head: | ||
return None | ||
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# keep looping until end of the links | ||
while current.next: | ||
current = current.next | ||
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return current.value | ||
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# method that inserts a given value as a new last node in the linked list | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(1) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ⏱ 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. |
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# Space Complexity: O(1) | ||
def add_last(self, value): | ||
pass | ||
current = self.head | ||
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if current is None: | ||
self.head = Node(value, None) | ||
else: | ||
while current: | ||
if current.next == None: | ||
break | ||
Comment on lines
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to
+108
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 🤓 We generally like to avoid break statements where possible. How might you refactor your code to eliminate this break statement? |
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current = current.next | ||
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current.next = Node(value, None) | ||
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# 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 | ||
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while current: | ||
if current.value > max: | ||
max = current.value | ||
current = current.next | ||
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return max | ||
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# 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 | ||
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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 | ||
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# 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 | ||
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@@ -86,24 +159,34 @@ def visit(self): | |
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# 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 | ||
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while current: | ||
next = current.next | ||
current.next = previous | ||
previous = current | ||
current = next | ||
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self.head = previous | ||
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## 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 | ||
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# 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 | ||
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# checks if the linked list has a cycle. A cycle exists if any node in the | ||
# linked list links to a node already visited. | ||
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@@ -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 | ||
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# navigate to last node | ||
current = self.head | ||
while current.next != None: | ||
current = current.next | ||
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current.next = self.head # make the last node link to first node | ||
pass |
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