C语言按规则合并为新链表

Question

运行没有输出，请大家帮我看看
题目：设链表 A=(a1,a2,…,am),B=(b1,b2,…,bn)，试编写一个按下列规则合并 A、B 为链表 C 的算法，使得: (利用原表结点空间)。
C=(a1,b1,…,am, bm, bm+1,…bn) m≤n
C=(b1,a1,…,bn, an, an+1,…am) m

#include <stdio.h>
#include <stdlib.h>
typedef struct node
{
    int date;
    struct node* next;
}node,*list;
list create(int*len);
int main()
{
    int m,n; 
    list A=create(&m);
    list B=create(&n);
    list C;
    list p=A->next,q=B->next;
    if(m<=n){
        C=A;
        while(p&&q){
            list t=p->next;
            list s=q->next;
            q->next=p->next;
            p->next=q;
            p=t;
            q=s;
        }
        p->next=q;
        free(B);
    }
    else if(m>n){
        while(p&&q){
        C=B;
        list t=q->next;
        list s=p->next;
        p->next=q->next;
        q->next=p;
        q=t;
        p=s;
        }
        q->next=p;
        free(A);
    }
    list f=C->next;
    while(f){
        printf("%d ",f->date);
    }
    return 0;
}
list create(int*len)
{
    int x,i=0;
    list head=(list)malloc(sizeof(node));
    head->date=0;
    head->next=NULL;
    list q=head;
    do{
        scanf("%d",&x);
        i++;
        list p=(list)malloc(sizeof(node));
        p->date=x;
        q->next=p;
        p->next=NULL;
        q=p;
    }while(getchar()!='\n');
    return head;
}

Answer 1

#include <stdio.h>
#include <stdlib.h>

typedef struct node
{
    int date;
    struct node *next;
} node, *list;

list create(int *len);

list insert(list P, list Q) // Q插入到P，保留P
{
    list p, q, t, s;
    p = P->next, q = Q->next;

    while (p->next)
    {
        t = p->next;
        s = q->next;
        p->next = q;
        q->next = t;
        p = t;
        q = s;
    }
    if (q)
        p->next = q;
    free(Q);
    return P;
}

int main()
{
    int m, n;
    list A = create(&m);
    list B = create(&n);
    list C;

    if (m <= n)
        C = insert(A, B);
    else
        C = insert(B, A);

    list f = C->next;

    while (f)
    {
        printf("%d ", f->date);
        f = f->next;
    }

    return 0;
}


list create(int *len)
{
    int x;//, i = 0;
    *len=0;
    list head = (list)malloc(sizeof(node));
    head->date = 0;
    head->next = NULL;
    list q = head;

    do
    {
        scanf("%d", &x);
        (*len)++;//i++;
        list p = (list)malloc(sizeof(node));
        p->date = x;
        q->next = p;
        p->next = NULL;
        q = p;
    } while (getchar() != '\n');
    
    return head;
}