C语言：合并有序双向链表问题

Question

从txt文件中读取5 4 2 3 1 10 6 8 7 9 到数组中，建立了两个递增排序的双向链表，内容分别为 1 2 3 4 5和6 7 8 9 10，现在想将两个链表合并输出一个递增的双向链表，但是！输出的链表少了1 和 6 两个数字，麻烦大佬看一下合并时哪里出了问题

#include "stdafx.h"
#include <stdlib.h>
#include <string.h>
#include <iostream>
using namespace std;
#include<fstream>
#include<stdio.h> 
/* C program to insetail nodes in doubly 
linked list such that list remains in 
ascending order on printing from left 
to right */

struct LinkList // A linked list node 
{ 
    int data; 
    struct LinkList *prior; 
    struct LinkList *next; 
}; 

// Function to insetail new node 
void nodeInsetail(struct LinkList **head, struct LinkList **tail, int key) 
{ 
    struct LinkList *p = new LinkList; 
    p->data = key; 
    p->next = NULL; 

    // If first node to be insetailed in doubly 
    // linked list 
    if (*head == NULL) 
    { 
        *head = p; 
        *tail = p; 
        (*head)->prior = NULL; 
        return; 
    } 
    // If node to be insetailed has value less 
    // than first node 
    if ((p->data) < ((*head)->data)) 
    { 
        p->prior = NULL; 
        (*head)->prior = p; 
        p->next = (*head); 
        (*head) = p; 
        return; 
    } 
    // If node to be insetailed has value more 
    // than last node 
    if ((p->data) > ((*tail)->data)) 
    { 
        p->prior = (*tail); 
        (*tail)->next = p; 
        (*tail) = p; 
        return; 
    } 
    // Find the node before which we need to 
    // insert p. 
    LinkList *temp = (*head)->next; 
    while ((temp->data) < (p->data)) 
        temp = temp->next; 

    // Insert new node before temp 
    (temp->prior)->next = p; 
    p->prior = temp->prior; 
    temp->prior = p; 
    p->next = temp; 
} 

// Function to print nodes in from left to right 
void printList(struct LinkList *temp) 
{ 
    while (temp != NULL) 
    { 
        printf("%d ", temp->data); 
        temp = temp->next; 
    } 
} 

// Driver program to test above functions 
int main() 
{ 
    int num[10],i,j=0;
    int datalen=0;
    ifstream file("linklist.txt");
    while( ! file.eof() )
    file>>num[datalen++];
    file.close();

    struct LinkList *La = NULL, *pa = NULL; 
    for(int i=0;i<5;i++)
    {
        nodeInsetail(&La, &pa,num[i]); 
    }
    printf("\nDoubly linked listA on printing:"); 
    printList(La);
    struct LinkList *Lb = NULL, *pb = NULL; 
    for(int i=5;i<10;i++)
    {
        nodeInsetail(&Lb, &pb,num[i]); 
    }
    printf("\nDoubly linked listB on printing:"); 
    printList(Lb);
    printf("\n");

    struct LinkList *Lc=NULL;//头
    struct LinkList *pc=NULL;//备用地址 
    struct LinkList *q=NULL;
    Lc=(struct LinkList*)malloc(sizeof(struct LinkList));
     pc=Lc;//备用合并链表起始地址 
     pa=La;
     pb=Lb;                    （这里有问题！）
     while(pa&&pb)
     {
        if(pa->data<pb->data)
        {
             Lc->next=pa;
             pa->prior=Lc;
             pa=pa->next;
             Lc=Lc->next;
        }
        else if(pa->data==pb->data)
        {
             Lc->next=pa;
             pa->prior=Lc;
             Lc=pa;
             pa=pa->next;
             q=pb->next;
             free(pb);
             pb=q;  
         }
        else if(pa->data>pb->data)
        {
             Lc->next=pb;
             pb->prior=Lc;
             pb=pb->next;
             Lc=Lc->next;
        }       
      }
     Lc->next=pa?pa:pb;
     free(Lb);
     printf("the result is:");
     pa=pc->next;
     while(pa)
     {
     printf("%d ",pa->data);
     pa=pa->next;
     }
    return 0;
}


执行结果如下（工作停止）
Doubly linked listA on printing:1 2 3 4 5
Doubly linked listB on printing:6 7 8 9 10
the result is:1 2 3 4 5 -572662307 请按任意键继续. . .

之前写的是pa=La->next;
pb=Lb->next; 总是少了1 和 6

Answer 1

是不是这里 如果链表没有头节点，归并排序初始化就不用指向头指针的next吧
(我没看完代码，如果不是这里问题请忽视)