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Print Linked List

Sar Champagne Bielert edited this page Apr 19, 2024 · 1 revision

Unit 5 Session 1 (Click for link to problem statements)

U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

  • How should the function handle an empty linked list (when head is None)?
    • The function should return an empty list if no nodes are present.

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the linked list starting from the head node, collect the values of each node, and return them in a list.

1) Initialize `current` as the starting node (`head`) of the linked list.
2) Create an empty list `values` to store the values of each node encountered.
3) Traverse through the linked list:
  a) Append the value of `current` to the `values` list.
  b) Move to the next node by updating `current` to `current.next`.
4) Continue the traversal until `current` is `None`.
5) Return the list `values` containing the values of all nodes in the linked list.

⚠️ Common Mistakes

  • Failing to check if the head is None, which could lead to attempts to access attributes of None.
  • Not updating current properly, leading to an infinite loop or missing nodes.

I-mplement

class Node:
    def __init__(self, value, next=None):
        self.value = value
        self.next = next
        
def print_linked_list(head):
    current = head
    values = []
    while current:
        values.append(current.value)
        current = current.next
    return values
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