ITPEC FE Subject B April 2026 Question 10

A is head1.val ≤ head2.val, B is head1.next and C is head2.next.

The merge takes the smaller of the two current heads and recurses on the rest:

• •Blank A — when head1.val is the smaller (or equal), head1 becomes the result node
• •Blank B — that node is now consumed, so the recursive call continues with head1.next while head2 stays where it is
• •Blank C — the mirror case: head2 was taken, so the call advances head2.next and leaves head1 untouched

Only the list whose element was just used may advance; the other must be reconsidered next time.

Answer (c)

Why not others:
- (a) and (b) — test head1.next is undefined, which asks about list length rather than order, so the output is not sorted
- (e) and (f) — take head1 only when the values are equal, sending every unequal pair down the head2 branch
- (g) and (h) — use ≥, which merges into descending order
- (b), (d), (f) and (h) — advance the list that was not consumed, dropping elements from one list and repeating them from the other

Key rule: In a recursive merge exactly one pointer moves per call, and it is the one whose value was just placed in the result.

AI-generated — may contain errors