Lilly C16 Pine #64
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Lilly C16 Pine #64
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spitsfire
reviewed
Jul 12, 2022
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(1)? | ||
| # Space Complexity: O(1)? | ||
| def get_first(self): |
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resubmitted with questions filled out
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| return self.head.value | ||
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| # method to add a new node with the specific data value in the linked list | ||
| # insert the new node at the beginning of the linked list | ||
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| def add_first(self, value): |
| pass | ||
| current = self.head | ||
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| while current != None: |
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We can shorten this to:
Suggested change
| while current != None: | |
| while current: |
| current = self.head | ||
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| count = 0 | ||
| while current != None: |
There was a problem hiding this comment.
Suggested change
| while current != None: | |
| while current: |
| if current.value == value: | ||
| return True | ||
| current = current.next | ||
| return False | ||
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| # method that returns the length of the singly linked list | ||
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| def length(self): |
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| while current is not None: | ||
| if current.value == value: | ||
| break | ||
| prev = current | ||
| current = current.next | ||
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| if current == None: | ||
| return None | ||
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| prev.next = current.next | ||
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| current = None |
There was a problem hiding this comment.
We can shorten this, too, to make it more readable:
Suggested change
| while current is not None: | |
| if current.value == value: | |
| break | |
| prev = current | |
| current = current.next | |
| if current == None: | |
| return None | |
| prev.next = current.next | |
| current = None | |
| while current is not None: | |
| if current.next.value == value: | |
| current.next = current.next.next | |
| current = current.next |
We don't need to re-assign current to None nor return None
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n)? | ||
| # Space Complexity: O(1)? | ||
| def reverse(self): |
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| current = self.head | ||
| count = 0 | ||
| while current: | ||
| count += 1 | ||
| current = current.next |
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hmm where have we seen this already? Maybe we can call length to get the number of nodes
| return current.value | ||
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| # checks if the linked list has a cycle. A cycle exists if any node in the | ||
| # linked list links to a node already visited. | ||
| # returns true if a cycle is found, false otherwise. | ||
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| def has_cycle(self): |
linked_list/linked_list.py
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| # 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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Resubmitted with time complexity |
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