The iterative version came naturally enough.
Start by clarifying the problem and edge cases, then walk through the iterative two-pointer approach with a dummy node, followed by the recursive approach. Analyze time and space complexity for both, emphasizing the trade-off between O(1) extra space for iterative and O(n+m) stack space for recursive. Finally, discuss handling duplicates and empty lists.
Pro tip: Mention that the iterative solution is generally preferred in production due to constant space, but the recursive solution showcases elegant code; also note that using a dummy node simplifies edge cases and reduces bugs.
Confirm that the lists are singly linked, sorted in ascending order, and that duplicates should be preserved. Discuss edge cases: one or both lists empty, lists of different lengths, and all elements in one list smaller than the other.
Explain using a dummy node to simplify list construction. Maintain a current pointer and compare nodes from both lists, appending the smaller one and advancing that list. After one list is exhausted, append the remainder of the other list.
Describe the recursive approach: if one list is empty, return the other; otherwise, compare the head nodes, set the smaller node's next to the recursive merge of the rest, and return the smaller node. Highlight base cases.
For both solutions, time complexity is O(n+m) where n and m are the lengths of the lists. Iterative space is O(1) extra space; recursive space is O(n+m) due to call stack. Mention that recursion depth could cause stack overflow for very long lists.
Compare iterative vs recursive: iterative is more space-efficient and avoids stack overflow, while recursive is more concise. Explain how duplicates are handled naturally by the comparison (using <= or < depending on stability). Confirm that empty lists are handled by returning the non-empty list.
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