Programming Ability Test学习 2-11. 两个有序链表序列的合并（15）

2-11. 两个有序链表序列的合并（15）

时间限制

500 ms

内存限制

80000 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的并集新非降序链表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：

在一行中输出合并后新的非降序链表，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：

序号	输入	输出
1	1 3 5 -1 2 4 6 8 10 -1	1 2 3 4 5 6 8 10
2	1 2 3 4 5 -1 1 2 3 4 5 -1	1 1 2 2 3 3 4 4 5 5
3	-1 -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;
    
    if(pthat->dex==-1&&pthis->dex==-1)printf("NULL\n");//两条都空 
    else if(pthat->dex==-1&&pthis->dex!=-1)//一条空输出另一条非空的 
    {
       while(pthis!=NULL&&pthis->Next->Next!=NULL){
       printf("%d",pthis->dex);
       if(pthis->Next->Next->Next!=NULL)printf(" ");
       else printf("\n");
       pthis=pthis->Next;
       }
    }
    else if(pthis->dex==-1&&pthat->dex!=-1)
    {
      while(pthat!=NULL&&pthat->Next->Next!=NULL){
       printf("%d",pthat->dex);
       if(pthat->Next->Next->Next!=NULL)printf(" ");
       else printf("\n");
       pthat=pthat->Next;
       }
    }
    else
    {
        struct Node *small=(pthat->dex>pthis->dex)?pthis:pthat;
        struct Node *big=(pthat->dex>pthis->dex)?pthat:pthis;
        struct Node *last=(pthat->dex>pthis->dex)?pthat:pthis;
        //printf("%d %d",small->dex,big->dex);
        pthis=small;
        pthat=big;
        while(pthis->dex<=pthat->dex&&pthis->dex!=-1&&pthat->dex!=-1)
        {
            last=pthis;
            pthis=pthis->Next;
            if(pthis->dex>=pthat->dex){
            struct Node *newNode=(node*)malloc(sizeof(node));
            newNode->dex=pthat->dex;
            newNode->Next=pthis;
            last->Next=newNode;
            pthis=last->Next;
            pthat=pthat->Next;
                 
              }
        }
           //如果samll链表用完了(small链表较短)
           if(pthat->dex!=-1)
           {
               last->Next=pthat;
           } 
        //遍历small 
        last=small;
        while(last->dex!=-1)
        {
            printf("%d",last->dex);
            if(last->Next->dex==-1)printf("\n");
            else printf(" ");
            last=last->Next; 
        }
        
    }
    
    //free(pthat);
    free(pthis);free(head);free(tail);free(head1);free(tail1);
    return 0;

}