What is the standard approach to delete all nodes containing a given value from a singly linked list, including edge cases?

• A. Use a dummy head; traverse and skip nodes with the target value by rewiring links; O(n)
• B. Scan and delete nodes by adjusting pointers without dummy head; O(n)
• C. Only delete first occurrence; O(n)
• D. Recompute list length after each deletion; O(n)

Answer: (B)

Using a dummy head is the cleanest way to handle deletions in a singly linked list, especially when the nodes at the front might match the target value. Create a placeholder node that points to the real head, then walk the list with a pointer that always sits at the node before the one being checked. If the next node’s value matches the target, skip it by rewiring the link (current.next = current.next.next); otherwise move forward. At the end, return the next of the dummy node as the new head. This approach naturally handles edge cases like all nodes matching, the head matching, or consecutive matching nodes, and it uses O(1) extra space with O(n) time.

Without a dummy head, you’d need extra logic to update the real head when it contains the target, and you must be careful with cases where the head changes multiple times. That makes the code more error-prone, even though it can be made to work.

Deleting only the first occurrence or recomputing length after each deletion don’t meet the goal of removing all matching nodes and add unnecessary work, respectively.