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Shaina Beth C16 Spruce#67

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shainabeth wants to merge 3 commits intoAda-C16:masterfrom
shainabeth:master
Open

Shaina Beth C16 Spruce#67
shainabeth wants to merge 3 commits intoAda-C16:masterfrom
shainabeth:master

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@shainabeth
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your space-time complexities look great! great job!

# Space Complexity: ?
# Time Complexity: O(1)
# Space Complexity: O(1)
def get_first(self):
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👍

# Space Complexity: ?
# Time Complexity: O(1)
# Space Complexity: O(1)
def add_first(self, value):
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👍

# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def search(self, value):
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👍

def search(self, value):
pass
cur = self.head
while cur != None:
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one way we can shorten this is by:

Suggested change
while cur != None:
while cur:

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

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

Comment on lines +91 to +95
prev = None
while cur != None:
prev = cur
cur = cur.next
prev.next = Node(value, None)
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same as above! we can start our while loop at while curnext ! None:

Suggested change
prev = None
while cur != None:
prev = cur
cur = cur.next
prev.next = Node(value, None)
while cur.next != None:
cur = cur.next
cur.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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👍

Comment on lines +122 to +129
prev = self.head
cur = self.head.next
while cur != None and cur.value != value:
prev = cur
cur = cur.next

if cur != None:
prev.next = cur.next
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this works! let's take away one of the variables, though. again, let's start the while loop with cur.next != None:

Suggested change
prev = self.head
cur = self.head.next
while cur != None and cur.value != value:
prev = cur
cur = cur.next
if cur != None:
prev.next = cur.next
cur = self.head
while cur.next != None and cur.next.value != value:
cur = cur.next
cur.next = cur.next.next

Comment on lines +151 to +156
cur = self.head
last_made = None
while cur != None:
last_made = Node(cur.value, last_made)
cur = cur.next
self.head = last_made
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hmm we shouldn't be adding nodes, only reversing the pointers so the last node is now the head and vice versa.

It's commonly done with three pointers: previous, current, and future node

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2 participants