Shaina Beth C16 Spruce #67
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spitsfire
reviewed
Jul 25, 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): |
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def search(self, value): |
| def search(self, value): | ||
| pass | ||
| cur = self.head | ||
| while cur != None: |
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one way we can shorten this is by:
Suggested change
| while cur != None: | |
| while cur: |
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def length(self): |
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def get_at_index(self, index): |
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| prev = None | ||
| while cur != None: | ||
| prev = cur | ||
| cur = cur.next | ||
| prev.next = Node(value, None) |
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same as above! we can start our while loop at while curnext ! None:
Suggested change
| prev = None | |
| while cur != None: | |
| prev = cur | |
| cur = cur.next | |
| prev.next = Node(value, None) | |
| while cur.next != None: | |
| cur = cur.next | |
| cur.next = Node(value, None) |
| prev = cur | ||
| cur = cur.next | ||
| prev.next = Node(value, None) | ||
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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): |
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| prev = self.head | ||
| cur = self.head.next | ||
| while cur != None and cur.value != value: | ||
| prev = cur | ||
| cur = cur.next | ||
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| if cur != None: | ||
| prev.next = cur.next |
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this works! let's take away one of the variables, though. again, let's start the while loop with cur.next != None:
Suggested change
| prev = self.head | |
| cur = self.head.next | |
| while cur != None and cur.value != value: | |
| prev = cur | |
| cur = cur.next | |
| if cur != None: | |
| prev.next = cur.next | |
| cur = self.head | |
| while cur.next != None and cur.next.value != value: | |
| cur = cur.next | |
| cur.next = cur.next.next |
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| cur = self.head | ||
| last_made = None | ||
| while cur != None: | ||
| last_made = Node(cur.value, last_made) | ||
| cur = cur.next | ||
| self.head = last_made |
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hmm we shouldn't be adding nodes, only reversing the pointers so the last node is now the head and vice versa.
It's commonly done with three pointers: previous, current, and future node
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