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Abigail C - Rock #55
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Abigail C - Rock #55
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Nice work Abigail, you hit the learning goals here. I left some minor feedback for you, but this was well done.
# Time Complexity: O(1) | ||
# Space Complexity: O(1) | ||
def get_first(self): |
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👍
# Time Complexity: O(1) | ||
# Space Complexity: O(1) | ||
def add_first(self, value): |
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👍
new_node = Node(value, next_node=self.head) | ||
self.head = new_node |
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Can be rewritten
new_node = Node(value, next_node=self.head) | |
self.head = new_node | |
self.head = Node(value, next_node=self.head) |
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
|
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def search(self, value): |
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👍
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
|
||
def length(self): |
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👍
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def delete(self, val): |
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👍
# Time Complexity: O(n) | ||
# Space Complexity: O(n) | ||
def visit(self): |
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👍
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def reverse(self): |
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👍
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def find_middle_value(self): |
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👍 This works, but you could either:
- use the
length
method - or have a fast and slow node reference where the fast move 2 nodes with each iteration and the slow moves one. Then when the fast reaches the end, the slow is referring to the middle node.
# # find the nth node from the end and return its value | ||
# # assume indexing starts at 0 while counting to n | ||
# # Time Complexity: O(n) | ||
# # Space Complexity: O(1) | ||
def find_nth_from_end(self, n): |
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👍 this works with similar feedback to find middle.
No description provided.