#include<stdio.h>
typedef struct node1{
    int data;
    struct node *next1;
}*L1list;
typedef struct node2{
    int data;
    struct node *next2;
}*L2list;
void print(L1list L1,L2list L2);
int judge(int m,L2list L2);
void result(L1list head1);
void intersec(L1list A,L2list B);
int main()
{

    int n,m,i;
    L1list p1,r1,q1,head1;
    L2list p2,r2,q2,head2;
    head1=NULL;
    head2=NULL;
    printf("输入第一个数据\n");
    scanf("%d",&n);
    while(n!=0)
    {
        p1=(L1list)malloc(sizeof(L1list));
        p1->data=n;
        p1->next1=NULL;
        if(head1==NULL)
        {
            head1=p1;
            r1=p1;
        }
        else
        {
            r1->next1=p1;
            r1=p1;
        }
        printf("输入下一个数\n");
        scanf("%d",&n);
    }
    getchar();
    printf("输入第二个数据\n");
    scanf("%d",&m);
    while(m!=0)
    {
        p2=(L2list)malloc(sizeof(L2list));
        p2->data=m;
        p2->next2=NULL;
        if(head2==NULL)
        {
            head2=p2;
            r2=p2;
        }
        else
        {
            r2->next2=p2;
            r2=p2;
        }
        printf("输入下一个数\n");
        scanf("%d",&m);
    }

 print(head1,head2);
 intersec(head1,head2);
 result(head1);
}
void result(L1list head1)
{
    L1list p;
    printf("A^B为\n");
    while(p)
    {
        printf("%d",p->data);
        p=p->next1;
    }
}

void print(L1list L1,L2list L2)
{
    L1list p1;
    L2list p2;
   
    p1=L1;
    p2=L2;
    printf("第一个链表为\n");
    while(p1)
    {
        printf("%d",p1->data);
        p1=p1->next1;
    }

    printf("第二个链表为\n");
        while(p2)
    {
        printf("%d",p2->data);
        p2=p2->next2;
    }
}
void intersec(L1list A,L2list B)
{
    L1list p,pr;
    int n;
    pr=A;
    p=pr;
    while(p)
    {
        n=p->data;
        if(!judge(n,B))
        {
          pr->next1=p->next1;
          free(p);
          while(p->next1)
          p=pr->next1;
        }
        else
        {
            pr=p;
            p=p->next1;
        }
    }
}
int judge(int m,L2list L2)
{
    L2list q;
    q=L2;
    if(q)
    {
        if(q->data==m)
        
        return 0;
          else
              q=q->next2;
     }
 else return 1;
}

            

和我看的...

[ 本帖最后由 Susake 于 2013-3-18 22:37 编辑 ]

不会吧，这是我自己原创的呀

不好意思，看快了...

编译不过，好多这种错误

cannot convert from 'struct node1 *' to 'struct node *'

细心点，慢慢改吧

这代码写的，有怎么复杂没，结构体都弄了2个，输入输出写的怎么繁琐

嘿嘿不懂