Merging two sorted linked lists

Question

This is one of the programming questions asked during written test from Microsoft. I am giving the question and the answer that I came up with. Thing is my answer although l

Answer 1

This is my take. Unlike other solutions it identifies and skips over consecutive nodes on one list which are smaller or equal to the head node of the other list. The head of the other list is attached at the end of such sequence and the process is repeated after switching roles. This approach minimizes the number of assignments to Node.next while limiting NULL testing to single check in each iteration.

Node * merge(Node *list1, Node *list2)
{
    if (!list1) return list2;
    if (!list2) return list1;

    Node *tmp;

    // compare head nodes and swap lists to guarantee list1 has the smallest node
    if (list1->val > list2->val) {
        tmp = list2;
        list2 = list1;
        list1 = tmp;
    }

    Node *tail = list1;

    do {
        // Advance the tail pointer skipping over all the elements in the result
        // which have smaller or equal value than the first node on list2
        while (tail->next && (tail->next->val <= list2->val)) {
            tail = tail->next;
        }
        // concat list2 at tail of result and make the rest after tail list2 
        tmp = tail->next;
        tail->next = list2;
        tail = list2;
        list2 = tmp;
    } while (list2);

    return list1;
}