What is the time complexity of reversing a singly linked list iteratively?

• A. O(1) time
• B. O(n log n) time
• C. O(n^2) time
• D. O(n) time

Answer: (D)

Analyzing time complexity by counting how many times we touch nodes. Reversing a singly linked list iteratively requires walking through each node once to flip its next pointer. In the loop, you typically use three pointers (prev, current, next) and perform a constant amount of work per node: save the next node, redirect current.next to point to the previous node, and move forward. Since each node is visited exactly one time, the total work grows linearly with the number of nodes, giving O(n) time.

This is faster than constant time, which would be impossible for a list of arbitrary length, and it doesn’t incur quadratic or n log n growth because there’s no nested loops or repeated passes—there’s just one pass with a fixed set of operations per node.