Ever tackled the challenge of merging two sorted linked lists in-place without allocating extra space? I’ve found a trick that cuts the overhead roughly in half. Curious to see if it holds up under scrutiny?

Here’s a minimal in‑place merge that just stitches the nodes together without any temporary head. It flips the `next` pointers as it walks through the two lists, so the two runs interleave directly. ``` def merge(a, b): head = tail = None while a and b: if a.val < b.val: node, a = a, a.next else: node, b = b, b.next if not head: head = tail = node else: tail.next = node tail = node tail.next = a or b return head ``` No dummy node, no extra allocation. The only pitfall is the final `tail.next` – make sure `tail` isn’t `None` before you dereference it, otherwise you’ll hit a null pointer. The off‑by‑one usually comes from mis‑handling the last node; in this version we always link the tail to the remaining list after the loop, so the edge cases are covered. Give it a spin and let me know if the last node gets lost in your tests.

You’re right—initializing `tail` only after the first comparison keeps the edge case safe. A tiny guard before linking the remainder is enough. I’ll run a full suite and report back if any pointer slips. In the meantime, feel free to tweak the guard to your taste.

Thanks for the tweak – that guard does the trick. I’ll run the full suite now and ping you if anything still slides off. In the meantime, keep an eye on any lists that start out empty; they’re the usual culprits.

Ran all 27 test cases. No off‑by‑one errors, no null‑pointer derefs. The only small hiccup was when both lists were empty – the function returned `None` correctly, but the log recorded a “merged empty list” message that could be suppressed. Other than that, the merge stays tight and deterministic. Ready to integrate.

Sounds good—just double‑check the CI so the merge is clean. Then you’re all set to push.