2-12. 两个有序链表序列的交集（20）

时间限制

400 ms

内存限制

64000 kB

代码长度限制

8000 B

判题程序

Standard

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。

输入格式说明：

输入分2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。

输出格式说明：

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”。

样例输入与输出：

1 2 5 -1
2 4 5 8 10 -1

2 5

1 3 5 -1
2 4 6 8 10 -1

NULL

1 2 3 4 5 -1
1 2 3 4 5 -1

1 2 3 4 5

3 5 7 -1
2 3 4 5 6 7 8 -1

3 5 7

-1
10 100 1000 -1

NULL

#include<stdio.h>
#include<stdlib.h>
#include <malloc.h>
typedef struct Node
{
   int dex;
   //int cof;
   struct Node * Next;
}node;

int main()
{
    //Qlink newLink;
    //newLink=(Link*)malloc(sizeof(Link));
    struct Node *head,*tail;
    head=(node*)malloc(sizeof(node));
    tail=(node*)malloc(sizeof(node));
    tail->Next=NULL;
    head->Next=tail;

    struct Node *pthis,*pthat;
    pthis=head;pthat=pthis;
    //第一个链表
    int dex;
    while(scanf("%d",&dex)){
    pthis=(node*)malloc(sizeof(node));
    pthis->dex=dex;
    pthis->Next=pthat->Next;
    pthat->Next=pthis;
    pthat=pthis;
    if(dex==-1)break;
    }
    //第二个链表
    struct Node *head1,*tail1;
    head1=(node*)malloc(sizeof(node));
    tail1=(node*)malloc(sizeof(node));
    tail1->Next=NULL;
    head1->Next=tail1;
    pthis=head1;pthat=pthis;
    getchar();
    while(scanf("%d",&dex)){
    pthis=(node*)malloc(sizeof(node));
    pthis->dex=dex;
    pthis->Next=pthat->Next;
    pthat->Next=pthis;
    pthat=pthis;
    if(dex==-1)break;
    }

    //遍历
    pthat=head->Next;
    pthis=head1->Next;

    //交集链表
        struct Node *head2,*tail2;
        head2=(node*)malloc(sizeof(node));
        tail2=(node*)malloc(sizeof(node));
        struct Node *newNode,*pnew;

        tail2->Next=NULL;
        head2->Next=tail2;
        newNode=head2;pnew=newNode;

     if(pthat->dex==-1||pthis->dex==-1)//存在空链表
       printf("NULL\n");
     else
     {

        while(pthis->dex!=-1&&pthat->dex!=-1)
        {
            if(pthis->dex==pthat->dex)//比较每一位
            {
               newNode=(node*)malloc(sizeof(node));
               newNode->dex=pthat->dex;
               newNode->Next=pnew->Next;
               pnew->Next=newNode;
               pnew=newNode;
               pthis=pthis->Next;
               pthat=pthat->Next;
            }
            else if(pthis->dex<pthat->dex)pthis=pthis->Next;
            else pthat=pthat->Next;

        }

        pthat=head2->Next;
        if(pthat->Next==NULL)printf("NULL\n");
        else{
        while(pthat->Next!=NULL)
        {
            printf("%d",pthat->dex);
            if(pthat->Next->Next==NULL)printf("\n");
            else printf(" ");
            pthat=pthat->Next;
        }
        }
     }

    //free(pthat);
    free(pthis);free(head);free(tail);free(head1);
    free(tail1);free(head2);free(tail2); free(newNode);
    return 0;

}