LeetCode 21: Merge Two Sorted Lists — Python Solution

LeetCode 21: Merge Two Sorted Lists is an Easy linked list problem. This Python walkthrough develops a dummy-tail merge solution, ties every code decision to a concrete invariant, and includes the regression check used before publication.

This independent guide paraphrases the task rather than reproducing LeetCode’s prompt. Use the official page for the exact statement, examples, constraints, and submission runner.

Recognizing the pattern

Linked-list solutions depend on preserving reachability while rewiring a constant number of pointers. Dummy nodes remove many head-only special cases.

For this problem specifically, attach the smaller current node to a dummy-backed output tail, advance that list, then attach the unexhausted suffix. The invariant worth writing beside the code is: The output through tail is sorted and contains exactly the consumed input nodes.

Step-by-step algorithm

1. Identify the input state consumed by mergeTwoLists(list1, list2) and initialize the data required by the dummy-tail merge pattern.
2. Attach the smaller current node to a dummy-backed output tail, advance that list, then attach the unexhausted suffix.
3. After each update, verify the page’s central invariant: The output through tail is sorted and contains exactly the consumed input nodes.
4. Finish only after the boundary behavior is covered: Either list may be empty; nodes can be reused rather than copied.

Python solution

LeetCode supplies the list, tree, or graph node class referenced by this method. The downloadable test suite includes compatible local node definitions so the implementation can also run outside the judge.

class Solution:
    def mergeTwoLists(self, list1, list2):
        dummy = tail = ListNode()
        while list1 and list2:
            if list1.val <= list2.val:
                tail.next, list1 = list1, list1.next
            else:
                tail.next, list2 = list2, list2.next
            tail = tail.next
        tail.next = list1 or list2
        return dummy.next

Reading the implementation

The main entry point is mergeTwoLists(list1, list2). The named working state includes dummy, tail; those variables make the dummy-tail merge state visible instead of hiding it in incidental control flow.

A single main loop advances the algorithm, which is the key reason the traversal does not revisit already resolved input unnecessarily. In concrete terms, attach the smaller current node to a dummy-backed output tail, advance that list, then attach the unexhausted suffix.

Correctness argument

Initialization. The data structure starts with exactly the information known before any input element is processed.

Preservation. Attach the smaller current node to a dummy-backed output tail, advance that list, then attach the unexhausted suffix. Each update records the current item without invalidating earlier decisions; consequently, the output through tail is sorted and contains exactly the consumed input nodes.

Termination. The traversal consumes a finite input or finite state space. At the end, the invariant covers the complete input, which is precisely the condition required for the returned result.

Complexity and trade-offs

O(m+n) time and O(1) extra space. The auxiliary-space figure excludes the returned output unless the output itself is the structure being built.

Copying values into an array makes indexing easy but abandons the intended pointer-space constraint and node identity. That comparison is useful in an interview because it explains why the final implementation is preferable, not merely that it passes.

Regression check

The published implementation belongs to a 100-problem suite that is compiled and exercised behaviorally before deployment.

One reference assertion from that suite is shown below. It targets the normal path while the edge conditions in the next section cover the failure-prone boundaries.

assert listed(Solution().mergeTwoLists(linked([1,2,4]),linked([1,3,4]))) == [1,1,2,3,4,4]

Common mistakes and edge cases

• Problem-specific boundary: Either list may be empty; nodes can be reused rather than copied.
• Pattern-level pitfall: Save the next node before changing a link, terminate reused tails, and check whether the original head can be removed.
• Invariant check: after every update, confirm that the output through tail is sorted and contains exactly the consumed input nodes.

Interview review checklist

• Explain why dummy-tail merge matches the structure of this input.
• State the invariant in one sentence before tracing code: The output through tail is sorted and contains exactly the consumed input nodes.
• Derive O(m+n) time and O(1) extra space from how many times each element or state is visited.
• Test the boundary explicitly: Either list may be empty; nodes can be reused rather than copied.

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