Interview Question: Merge two sorted singly linked lists without creating new nodes

Question

This is a programming question asked during a written test for an interview. \"You have two singly linked lists that are already sorted, you have to merge them and return a

Answer 1

My take on the question is as below:

Pseudocode:

Compare the two heads A and B. 
If A <= B, then add A and move the head of A to the next node. 
Similarly, if B < A, then add B and move the head of B to the next node B.
If both A and B are NULL then stop and return.
If either of them is NULL, then traverse the non null head till it becomes NULL.

Code:

public Node mergeLists(Node headA, Node headB) {
    Node merge = null;
    // If we have reached the end, then stop.
    while (headA != null || headB != null) {
        // if B is null then keep appending A, else check if value of A is lesser or equal than B
        if (headB == null || (headA != null && headA.data <= headB.data)) {
            // Add the new node, handle addition separately in a new method.
            merge = add(merge, headA.data);
            // Since A is <= B, Move head of A to next node
            headA = headA.next;
        // if A is null then keep appending B, else check if value of B is lesser than A
        } else if (headA == null || (headB != null && headB.data < headA.data)) {
            // Add the new node, handle addition separately in a new method.
            merge = add(merge, headB.data);
            // Since B is < A, Move head of B to next node
            headB = headB.next;
        }
    }
    return merge;
}

public Node add(Node head, int data) {
    Node end = new Node(data);
    if (head == null) {
        return end;
    }

    Node curr = head;
    while (curr.next != null) {
        curr = curr.next;
    }

    curr.next = end;
    return head;
}