Spruce - Angela Fan #71
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added 2 commits
July 17, 2022 15:38
kyra-patton
reviewed
Jul 28, 2022
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(1) | ||
| # Space Complexity: O(1) | ||
| def get_first(self): |
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(1) | ||
| # Space Complexity: O(1) | ||
| def add_first(self, 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(1) |
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⏱ Time complexity is O(n) since in the worst case the node with the value you are searching for will be the last node in the list, and you will have to iterate through every node in the list.
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(1) | ||
| # Space Complexity: O(1) | ||
| def search(self, value): |
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def length(self): |
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| # Time Complexity: ? | ||
| # Space Complexity: ? |
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| # Time Complexity: ? | ||
| # Space Complexity: ? |
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| # Time Complexity: ? | ||
| # Space Complexity: ? |
| previous = current | ||
| current = next_node | ||
| self.head = previous | ||
| return 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): |
| current = current.next | ||
| middle_value = current.value | ||
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| return middle_value | ||
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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): |
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You could also take advantage of the get_at_index method you wrote
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