How do you merge two sorted singly linked lists into a new sorted list using a dummy head and existing nodes?

• A. Allocate a brand-new list by creating new nodes for every value; O(m+n) time, O(m+n) space.
• B. Reverse one list, then merge by alternating nodes.
• C. Use a dummy head, repeatedly attach the smaller of the two list heads to the merged list, advance that list; append the remainder; O(m+n) time, O(1) extra space.
• D. If the lists are already sorted, simply concatenate them.

Answer: (C)

Merging two sorted singly linked lists with a dummy head shows how to knit the lists together in place while preserving order. A dummy head acts as a fixed starting point, so you don’t have to special-case the first node. You keep two pointers to the current nodes of the input lists and a tail pointer for the merged list. At each step, attach the smaller current node to the tail of the merged list, then advance the pointer that you took the node from. Move the tail forward as you go. When one list runs out, append the rest of the other list, since it’s already sorted. Because you only walk through each list once, the time is O(m+n). You reuse the existing nodes, so the extra space is O(1) aside from the dummy node. This approach also handles empty input lists gracefully.

Creating new nodes would require extra space proportional to the total number of elements. Reversing one list and alternating nodes would disrupt the sorted order in general, and simply concatenating the lists only yields a sorted result if all elements of the first list are <= all elements of the second.