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kyra-patton
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Jul 28, 2022
| # returns None if the list is empty | ||
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: Linear O(N), size of LL |
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⏱ Time complexity would actually be O(1) here, because you only ever need to look at the head of the list.
| # Space Complexity: ? | ||
| # Time Complexity: Adding a head, Constant O(1) | ||
| # Space Complexity: Constant O(1) not creating new data structure | ||
| def add_first(self, value): |
| # Space Complexity: ? | ||
| # Time Complexity: Linear O(N), size of the LL | ||
| # Space Complexity: Constant O(1), we're not creating data structures | ||
| def search(self, value): |
| # Space Complexity: ? | ||
| # Time Complexity: Linear O(N), size of the LL | ||
| # Space Complexity: Constant O(1), we're not creating data structures | ||
| def length(self): |
| # Space Complexity: ? | ||
| # Time Complexity: Linear O(N), size of the LL | ||
| # Space Complexity: Constant O(1), we're not creating data structures | ||
| def get_at_index(self, index): |
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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): |
| # Space Complexity: ? | ||
| # Time Complexity: Linear O(N), size of the LL | ||
| # Space Complexity: Constant O(1), we're not creating data structures | ||
| def delete(self, value): |
| # Space Complexity: ? | ||
| # Time Complexity: Linear O(N), size of the LL | ||
| # Space Complexity: Linear O(N), depending on the size of the visited_list | ||
| def visit(self): |
| visited_list = [] | ||
| current = self.head | ||
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| while current: |
| # Space Complexity: ? | ||
| # Time Complexity: Linear O(N), depending on the size of the LL | ||
| # Space Complexity: Constant O(1), no creation of new data structures | ||
| def reverse(self): |
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