Caballero- SPRUCE #61
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| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(1) accessing it only on | ||
| # Space Complexity: O(1) no new data structure being created | ||
| def get_first(self): |
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: o(1) | ||
| # Space Complexity: o(1) constant time, since new data structure is being created but its small | ||
| def add_first(self, value): |
| pass | ||
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| self.head = Node(value,self.head) | ||
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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: ? | ||
| def search(self, value): |
| # method to find if the linked list contains a node with specified value | ||
| # returns true if found, false otherwise | ||
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| def search(self, value): | ||
| pass | ||
| current = self.head | ||
| while current is not None: |
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we can shorten it to:
| while current is not None: | |
| while current: |
| # Time Complexity: ? | ||
| # Space Complexity: ? |
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What do you think the space and time complexity of this is? Is the time complexity dependent on the length of the linked list? Is there any new variables being created that grow larger depending upon the input?
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| # method that returns the length of the singly linked list | ||
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: 0(1) constant time |
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This should be O(n). The while loop is dependent upon the length of the linked list (ie, the input) to find the appropriate node
| length += 1 | ||
| current = current.next | ||
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| return length |
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Move this line out of the while loop; otherwise, it will return immediately after the first iteration:
| return length | |
| return length |
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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: ? | ||
| def get_at_index(self, index): | ||
| pass | ||
| if self.length() < index: |
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love this! use those helper methods!
| length += 1 | ||
| current = current.next | ||
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| return length | ||
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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: ? | ||
| def get_at_index(self, index): |
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great start, but if the index is less than the length of the linked list, how do we find the appropriate node value at that index?
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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: ? | ||
| def get_at_index(self, index): | ||
| pass | ||
| if self.length() < index: | ||
| 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: ? | ||
| def get_last(self): |
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How do we travel to the down to the last node and return its value?
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