链表E 有序链表的归并

Problem Description
分别输入两个有序的整数序列（分别包含M和N个数据），建立两个有序的单链表，将这两个有序单链表合并成为一个大的有序单链表，并依次输出合并后的单链表数据。
Input
第一行输入M与N的值；
第二行依次输入M个有序的整数；
第三行依次输入N个有序的整数。
Output
输出合并后的单链表所包含的M+N个有序的整数。
Example Input
6 5
1 23 26 45 66 99
14 21 28 50 100
Example Output
1 14 21 23 26 28 45 50 66 99 100

#include <stdio.h>#include <stdlib.h>struct node{    int data;    struct node *next;};struct node *creat(int t)   //顺序建立链表{    int i;    struct node *head,*p,*tail;    head = (struct node*)malloc(sizeof(struct node));    head->next = NULL;    tail = head;    for(i=0; i<t; i++)    {        p = (struct node*)malloc(sizeof(struct node));        scanf("%d",&p->data);        p->next = NULL;        p->next = tail->next;        tail->next = p;        tail = p;    }    return head;};struct node *merge(struct node *head1,struct node *head2)   //归并两个链表{    struct node *p1,*p2,*tail;    p1 = head1->next;    p2 = head2->next;    tail = head1;    free(head2);    while(p1!=NULL&&p2!=NULL)    {        if(p1->data<p2->data)        {            tail->next = p1;            tail = p1;            p1 = p1->next;        }        else        {            tail->next = p2;            tail = p2;            p2 = p2->next;        }    }    if(p1!=NULL)    {        tail->next = p1;    }    else    {        tail->next = p2;    }    return(head1);};void show(struct node *head)   //显示链表{    struct node *p;    p = head->next;    printf("%d",p->data);    p = p->next;    while(p!=NULL)  //不知个数用while条件    {        printf(" %d",p->data);        p = p->next;    }    printf("\n");};int main(){    int m,n;    struct node *head1,*head2,*head;    scanf("%d %d",&m,&n);    head1 = creat(m);    head2 = creat(n);    head = merge(head1,head2);    show(head);    return 0;}