London | 25-SDC-July | Andrei Filippov | Sprint 2 | Implement a linked list in Python #28
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Droid-An
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cjyuan
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Jan 25, 2026
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| def push_head(self, element): | ||
| node = Node(element) | ||
| if self.head and self.tail: |
There was a problem hiding this comment.
Is it necessary to check also self.tail?
| node.next = self.head | ||
| self.head.previous = node | ||
| self.head = node | ||
| elif not self.head and not self.tail: |
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This line could probably be simplified.
Comment on lines
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| if self.head != node and self.tail != node: | ||
| node.previous.next = node.next | ||
| node.next.previous = node.previous | ||
| node.previous, node.next = None, None | ||
| elif self.head == node and self.tail != node: | ||
| node.next.previous = None | ||
| self.head = node.next | ||
| node.previous, node.next = None, None | ||
| elif self.head != node and self.tail == node: | ||
| node.previous.next = None | ||
| self.tail = node.previous | ||
| node.previous, node.next = None, None | ||
| elif self.head == node and self.tail == node: |
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The conditions can be simplified (reduce the amount of checking needed) as:
if self.head == node:
if self.tail == node:
# Case 1: head == node, tail == node
else:
# Case 2: head == node, tail != node
else;
if self.tail == node:
# Case 3: head != node, tail == node
else:
# Case 4: head != node, tail != nodeOnly two comparisons needed.
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Learners, PR Template
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Changelist
Implemented a linked list in Python.
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