Explain how to implement a basic iterator for a singly linked list.

• A. Maintain a current pointer; hasNext checks current != NULL; next returns current.data and advances current to current.next.
• B. Maintain a pointer to head and always reset on hasNext.
• C. Use a stack to simulate iteration.
• D. Use a for loop with indexing i from 0 to length-1.

Answer: (A)

The fundamental idea is to keep a moving pointer to the node you’re about to read and advance that pointer as you go. When an iterator starts, it points to the head of the list. hasNext simply checks whether that current pointer is not null, meaning there is a node left to visit. When you call next, you return the data from the current node and then advance the pointer to the next node in the list. This approach matches the structure of a singly linked list, where each node only knows its successor, so you walk through the list by following next pointers and you don’t have random access to elements by index.

This design keeps the iteration efficient: each next operation is O(1), and traversing the entire list is O(n) time with O(1) extra space for the iterator. For an empty list, the current pointer starts as null, so hasNext is false from the outset and next shouldn’t be called.

Other approaches introduce unnecessary complications. Resetting on hasNext would mutate the iterator’s progress during a check, which is undesirable. Using a stack would add extra memory and is overkill for a straightforward forward iteration (it’s sometimes used for specific traversal orders or reverse iteration). Trying to index with a for loop is not practical on a singly linked list because there’s no constant-time access to the i-th element; you'd have to walk from the head each time, making the process inefficient.