Which sequence correctly reverses a singly linked list iteratively?

• A. Initialize head to null and iterate, creating a reversed list.
• B. Initialize prev to null and curr to head; while curr: next = curr.next; curr.next = prev; prev = curr; curr = next; head = prev.
• C. Start from head and swap data in nodes to reverse order.
• D. Reverse by iterating to the end and then moving pointers backwards.

Answer: (B)

Reversing a singly linked list iteratively means flipping the next pointers so every node points to its predecessor, not its successor, and doing this in place as you traverse once. The standard in-place method uses three pointers: start with prev as null and curr at the head. While curr exists, save the next node (so you can continue later), set curr.next to prev to reverse the link, move prev forward to curr, and advance curr to the saved next. When the loop finishes, prev is the new head, so you update head to prev. This exact sequence performs the pointer reversal cleanly and in constant extra space, without creating new nodes or just swapping data. The other options either avoid changing the actual links, attempt backwards movement without extra structure, or create a new reversed list rather than reversing in place.