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

Answer 2

 
#include <iostream>
using namespace std;
int main()
{
    int a[32768], b[32768], A[32768] = {0}, c, d, e = 0;
    cin >> c;
    for (int i = 0; i < c; i++)
{
        cin >> a[i];
        A[a[i]] = 1;
}
    cin >> d;
    for (int i = 0; i < d; i++)
{
        cin >> b[i];
}
    for (int i = 0; i < d; i++)
{
        if (A[b[i]] == 1)
{
            if (e != 0)
{
                cout << " ";
}
e++;
            cout << b[i];
            A[b[i]] = 0; // 标记该数字已输出
}
}
    if (e == 0)
{
        cout << "0";
}
 
    return 0;
}
 
 

Answer 3

亲手写的，望采纳

Answer 4

蟹蟹，刚学链表，所以用链表来练习