这输出老为空是怎个事啊

Question

已知两个非降序链表序列S1与S2，设计函数构造出S1与S2的交集新链表S3。
输入格式:
输入分两行，分别在每行给出由若干个正整数构成的非降序序列，用−1表示序列的结尾（−1不属于这个序列）。数字用空格间隔。
输出格式:
在一行中输出两个输入序列的交集序列，数字间用空格分开，结尾不能有多余空格；若新链表为空，输出NULL。
输入样例:
1 2 5 -1
2 4 5 8 10 -1
输出样例:
2 5

#include <iostream>

using namespace std;

typedef struct Lnode
{
    int data;
    Lnode *next;
}Lnode, *Linklist;

void creat_a( Linklist &La )
{
    Linklist p, q;
    La = new Lnode;
    La->next = NULL;
    q = La;
    while (1)
    {
        p = new Lnode;
        cin>>p->data;
        if ( p->data == -1 )
        {
            return;
        }
        p->next = NULL;
        q->next = p;
        q = p;
    }
}

void creat_b ( Linklist &Lb )
{
    Linklist p, q;
    Lb = new Lnode;
    Lb->next = NULL;
    q = Lb;
    while(1)
    {
        p = new Lnode;
        cin>>p->data;
        if ( p->data == -1 )
        {
            return;
        }
        p->next = NULL;
        q->next = p;
        q = p;
    }
}

void jiaoji ( Linklist &La, Linklist &Lb, Linklist &Lc )
{
    Linklist pa, pb, q, r;
    Lc = new Lnode;
    Lc->next = NULL;
    pa = La;
    pb = Lb;
    r = Lc;
    while ( pa->next != NULL )
    {
        while ( pb->next != NULL )
        {
            if ( pa->next->data == pb->next->data )
            {
                q = new Lnode;
                q->data = pb->next->data;
                q->next = NULL;
                r->next = q;
                r = q;
            }
            pb = pb->next;
        }
        pa = pa->next;
    }
}

void Print(Linklist Lc)
{
    Linklist p = Lc->next;
    if ( p == NULL )
    {
        cout << "NULL";
    }
    while (p != NULL)
    {
        cout << p->data;
        if (p->next != NULL)
        {
            cout << " ";
        }
        p = p->next;
    }
}
int main ()
{
    Linklist La, Lb, Lc;
    creat_a (La);
    creat_b(Lb);
    jiaoji (La, Lb, Lc);
    Print (Lc);
    return 0;
}

Answer 1

这个是找交集的程序吗，我有更简单的