How to merge sort already sorted linked lists?

Question

I'm writing a method that merges two already sorted linked lists. However, the very last node of the list won't print out for some reason. Any ideas?

#include <iostream>
using namespace std;

struct node {
        int number;
        node *next;
        node *prev;
};

    void mergeSort(node*& head, node*& head1);

    int main(int argc, const char * argv[])
    {

//Linked List #1 
        node *head;
        node *tail;
        node *curr;

//1st node
        curr = new node;
        curr-> number = 1;
        curr-> prev = NULL;
        head = curr;
        tail = curr;

        //2nd node
        curr = new node;
        curr-> number = 3;
        curr -> prev = tail;
        tail -> next = curr;
        tail = curr;

        //3rd node
        curr = new node;
        curr-> number = 5;
        curr -> prev = tail;
        tail -> next = curr;
        tail = curr;

        //4th node (tail)
        curr = new node;
        curr-> number = 7;
        curr -> prev = tail;
        tail -> next = curr;
        tail = curr;
        tail->next = NULL;

        //Print Linked List #1
        cout<< "Linked List #1: " << endl;
        curr = head;
        while (curr != NULL){
            cout << curr-> number;
            curr = curr->next;
        }
        cout << endl;

    //Linked List #2
        node *head1;
        node *tail1;
        node *curr1;

        //1st node
        curr1 = new node;
        curr1-> number = 2;
        curr1-> prev = NULL;
        head1 = curr1;
        tail1 = curr1;

        //2nd node
        curr1 = new node;
        curr1-> number = 4;
        curr1 -> prev = tail1;
        tail1 -> next = curr1;
        tail1 = curr1;

        //3rd node
        curr1 = new node;
        curr1-> number = 6;
        curr1 -> prev = tail1;
        tail1 -> next = curr1;
        tail1 = curr1;

        //4th node (tail)
        curr1 = new node;
        curr1-> number = 8;
        curr1 -> prev = tail1;
        tail1 -> next = curr1;
        tail1 = curr1;
        tail1->next = NULL;

        //Print Linked List #2
        cout<< "Linked List #2: " << endl;
        curr1 = head1;
        while (curr1 != NULL){
            cout << curr1-> number;
            curr1 = curr1->next;
        }
        cout << endl;

        //Call MergeSort function
        mergeSort(head, head1);


        return 0;
    }

Here is the merge sort method for the Linked Lists.

    void mergeSort(node*& head, node*& head1){

        //Set up the Merge Sorted Linked List

        node *head2;
        node *tail2;
        node *curr2;

        node *curr = head;
        node *curr1 = head1;

        node* next = curr->next;
        node* next1 = curr1->next;
        node* next2 = curr2->next;

        //1st NODE
        curr2 = new node;

        if (curr->number > curr1->number){
            curr2->number = curr1->number;
            curr1 = curr1->next;
            curr1 = next1;
        }
        else if (curr1->number > curr->number){
            curr2->number = curr->number;
            curr = curr->next;
            curr = next;
        }


        //Set prev of head to Null
        curr2 -> prev = NULL;
        //Set head
        head2 = curr2;
        //Set tail
        tail2 = curr2;



        //BODY NODES
        while (curr != NULL && curr1 != NULL ){
            curr2 = new node;



        //compare the data between the two nodes

            //compare the data between the two nodes
            if (curr1->number >= curr -> number){
                //set the new node's data to the smallest of the previous 
              //node's datas
                //from the other two Linked Lists
                curr2 -> number = curr -> number;

                //link the new node to the previous
                curr2 -> prev = tail2;
                //attach the predecessor node's next pointer to the current         
                //node
                tail2 -> next = curr2;
                //set the new node as the tail
                tail2 = curr2;

                //iterate through the selected Linked List
                curr = curr -> next;



            }


            else if (curr->number >= curr1-> number){
           //set the new node's data to the smallest of the previous node's 
           //datas
            //from the other two Linked Lists
                curr2 -> number = curr1 -> number;

            //link the new node to the previous
                curr2 -> prev = tail2;
            //attach the predecessor node's next pointer to the current node
                tail2 -> next = curr2;
            //set the new node as the tail
                tail2 = curr2;

            //iterate through the selected Linked List
                curr1 = curr1 -> next;

            }


        } tail2 -> next = NULL;



    //Print Linked List #3
            cout<< "Linked List #3: " << endl;
            curr2 = head2;
            while (curr2){
                cout << curr2-> number;
                curr2 = curr2->next;

            }
            cout << endl;

    }

Answer 1

The merge function should not be allocating nodes, just linking the existing node's pointers. Here's an example to merge two single linked lists. If you need previous pointers, that could be done after doing the merge:

NODE * MergeLists(NODE *pSrc1, NODE *pSrc2)
{
NODE *pDst = NULL;                      /* destination head ptr */
NODE **ppDst = &pDst;                   /* ptr to head or prev->next */
    while(1){
        if(pSrc1 == NULL){
            *ppDst = pSrc2;
            break;
        }
        if(pSrc2 == NULL){
            *ppDst = pSrc1;
            break;
        }
        if(pSrc2->data < pSrc1->data){  /* if src2 < src1 */
            *ppDst = pSrc2;
            pSrc2 = *(ppDst = &(pSrc2->next));
            continue;
        } else {                        /* src1 <= src2 */
            *ppDst = pSrc1;
            pSrc1 = *(ppDst = &(pSrc1->next));
            continue;
        }
    }
    return pDst;
}