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

void printLL(){
    NodeLL cur = head;
    if(cur.getNext() == null){
        System.out.println("LL is emplty");
    }else{
        //System.out.println("printing Node");
        while(cur.getNext() != null){
            cur = cur.getNext();
            System.out.print(cur.getData()+ " ");

        }
    }
    System.out.println();
}

void mergeSortedList(NodeLL node1, NodeLL node2){
    NodeLL cur1 = node1.getNext();
    NodeLL cur2 = node2.getNext();

    NodeLL cur = head;
    if(cur1 == null){
        cur = node2;
    }

    if(cur2 == null){
        cur = node1;
    }       
    while(cur1 != null && cur2 != null){

        if(cur1.getData() <= cur2.getData()){
            cur.setNext(cur1);
            cur1 = cur1.getNext();
        }
        else{
            cur.setNext(cur2);
            cur2 = cur2.getNext();
        }
        cur = cur.getNext();
    }       
    while(cur1 != null){
        cur.setNext(cur1);
        cur1 = cur1.getNext();
        cur = cur.getNext();
    }       
    while(cur2 != null){
        cur.setNext(cur2);
        cur2 = cur2.getNext();
        cur = cur.getNext();
    }       
    printLL();      
}

Answer 2

Here is a complete working example that uses the linked list implemented java.util. You can just copy paste the code below inside a main() method.

        LinkedList<Integer> dList1 = new LinkedList<Integer>();
        LinkedList<Integer> dList2 = new LinkedList<Integer>();
        LinkedList<Integer> dListMerged = new LinkedList<Integer>();

        dList1.addLast(1);
        dList1.addLast(8);
        dList1.addLast(12);
        dList1.addLast(15);
        dList1.addLast(85);

        dList2.addLast(2);
        dList2.addLast(3);
        dList2.addLast(12);
        dList2.addLast(24);
        dList2.addLast(85);
        dList2.addLast(185);

        int i = 0;
        int y = 0;
        int dList1Size = dList1.size();
        int dList2Size = dList2.size();
        int list1Item = dList1.get(i);
        int list2Item = dList2.get(y);
        while (i < dList1Size || y < dList2Size) {

            if (i < dList1Size) {

                if (list1Item <= list2Item || y >= dList2Size) {
                    dListMerged.addLast(list1Item);
                    i++;
                    if (i < dList1Size) {
                        list1Item = dList1.get(i);
                    }
                }
            }


            if (y < dList2Size) {

                if (list2Item <= list1Item || i >= dList1Size) {
                    dListMerged.addLast(list2Item);
                    y++;
                    if (y < dList2Size) {
                        list2Item = dList2.get(y);
                    }
                }
            }

        }

        for(int x:dListMerged)
        {
            System.out.println(x);
        }

Answer 3

Here is the algorithm on how to merge two sorted linked lists A and B:

while A not empty or B not empty:
   if first element of A < first element of B:
      remove first element from A
      insert element into C
   end if
   else:
      remove first element from B
      insert element into C
end while

Here C will be the output list.

Answer 4

Iteration can be done as below. Complexity = O(n)

public static LLNode mergeSortedListIteration(LLNode nodeA, LLNode nodeB) {
    LLNode mergedNode ;
    LLNode tempNode ;      

    if (nodeA == null) {
        return nodeB;
      } 
      if (nodeB == null) {
        return nodeA;
      }     


    if ( nodeA.getData() < nodeB.getData())
    {
        mergedNode = nodeA;
        nodeA = nodeA.getNext();
    }
    else
    {
        mergedNode = nodeB;
        nodeB = nodeB.getNext();
    }

    tempNode = mergedNode; 

    while (nodeA != null && nodeB != null)
    {           

        if ( nodeA.getData() < nodeB.getData())
        {               
            mergedNode.setNext(nodeA);
            nodeA = nodeA.getNext();
        }
        else
        {
            mergedNode.setNext(nodeB);
            nodeB = nodeB.getNext();                
        }       
        mergedNode = mergedNode.getNext();
    }

    if (nodeA != null)
    {
        mergedNode.setNext(nodeA);
    }

    if (nodeB != null)
    {
        mergedNode.setNext(nodeB);
    }       
    return tempNode;
}

Answer 5

        /* Simple/Elegant Iterative approach in Java*/    
        private static LinkedList mergeLists(LinkedList list1, LinkedList list2) {
                    Node head1 = list1.start;
                    Node head2 = list2.start;
                    if (list1.size == 0)
                    return list2;
                    if (list2.size == 0)
                    return list1;               
                    LinkedList mergeList = new LinkedList();
                    while (head1 != null && head2 != null) {
                        if (head1.getData() < head2.getData()) {
                            int data = head1.getData();
                            mergeList.insert(data);
                            head1 = head1.getNext();
                        } else {
                            int data = head2.getData();
                            mergeList.insert(data);
                            head2 = head2.getNext();
                        }
                    }
                    while (head1 != null) {
                        int data = head1.getData();
                        mergeList.insert(data);
                        head1 = head1.getNext();
                    }
                    while (head2 != null) {
                        int data = head2.getData();
                        mergeList.insert(data);
                        head2 = head2.getNext();
                    }
                    return mergeList;
                }

/* Build-In singly LinkedList class in Java*/
class LinkedList {
    Node start;
    int size = 0;

    void insert(int data) {
        if (start == null)
            start = new Node(data);
        else {
            Node temp = start;
            while (temp.getNext() != null) {
                temp = temp.getNext();
            }
            temp.setNext(new Node(data));
        }
        size++;
    }

    @Override
    public String toString() {

        String str = "";
        Node temp=start;
        while (temp != null) {
            str += temp.getData() + "-->";
            temp = temp.getNext();
        }
        return str;
    }

}

Answer 6

// Common code for insert at the end
        private void insertEnd(int data) {
                Node newNode = new Node(data);
                if (head == null) {
                    newNode.next = head;
                    head = tail = newNode;
                    return;
                }
                Node tempNode = tail;
                tempNode.next = newNode;
                tail = newNode;
            }

    private void mergerTwoSortedListInAscOrder(Node tempNode1, Node tempNode2) {

            if (tempNode1 == null && tempNode2 == null)
                return;
            if (tempNode1 == null) {
                head3 = tempNode2;
                return;
            }
            if (tempNode2 == null) {
                head3 = tempNode1;
                return;
            }

            while (tempNode1 != null && tempNode2 != null) {

                if (tempNode1.mData < tempNode2.mData) {
                    insertEndForHead3(tempNode1.mData);
                    tempNode1 = tempNode1.next;
                } else if (tempNode1.mData > tempNode2.mData) {
                    insertEndForHead3(tempNode2.mData);
                    tempNode2 = tempNode2.next;
                } else {
                    insertEndForHead3(tempNode1.mData);
                    insertEndForHead3(tempNode2.mData);
                    tempNode1 = tempNode1.next;
                    tempNode2 = tempNode2.next;
                }

            }
            if (tempNode1 != null) {
                while (tempNode1 != null) {
                    insertEndForHead3(tempNode1.mData);
                    tempNode1 = tempNode1.next;
                }
            }
            if (tempNode2 != null) {
                while (tempNode2 != null) {
                    insertEndForHead3(tempNode2.mData);
                    tempNode2 = tempNode2.next;
                }
            }
        }

:)GlbMP