Question

数据结构课程设计----编制一个能演示执行集合的并交和差运算的程序。（ 用有序链表表示集合）

1.集合的元素限定为小写字母字符【'a'..'z'】。
2.演示程序以用户和计算机的对话方式执行。

Answer 1

#include<stdio.h>
#include<stdlib.h>

typedef struct pointer
{
char data;
struct pointer *link;
} pointer;

void readdata(pointer *head){ //读集合
pointer *p;
char tmp;
printf("input data ('0' for end):");
scanf("%c",&tmp);
while(tmp!='0')
{
if((tmp<'a')||(tmp>'z'))
{
printf("输入错误！必须为小写字母！\n");
return;
}
p=(pointer *)malloc(sizeof(struct pointer));
p->dat=tmp;
p->link=head->link;
head->link=p;
scanf("%c",&tmp);
}
}

void disp(pointer *head){ //显示集合数据
pointer *p;
p=head->link;
while(p!=NULL)
{
printf("%c ",p->dat);
p=p->link;
}
printf("\n");
}

void bing(pointer *head1,pointer *head2, pointer *head3){ //计算集合1与集合2的并
pointer *p1,*p2,*p3;
p1=head1->link;
while(p1!=NULL)
{
p3=(pointer *)malloc(sizeof(struct pointer));
p3->dat=p1->dat;
p3->link=head3->link;
head3->link=p3;
p1=p1->link;
}
p2=head2->link;
while(p2!=NULL)
{
p1=head1->link;
while((p1!=NULL)&&(p1->dat!=p2->dat))
p1=p1->link;
if(p1==NULL)
{
p3=(pointer *)malloc(sizeof(struct pointer));
p3->dat=p2->dat;
p3->link=head3->link;
head3->link=p3;
}
p2=p2->link;
}
}

void jiao(pointer *head1,pointer *head2, pointer *head3){ //计算集合1与集合2的交
pointer *p1,*p2,*p3;
p1=head1->link;
while(p1!=NULL)
{
p2=head2->link;
while((p2!=NULL)&&(p2->dat!=p1->dat))
p2=p2->link;
if((p2!=NULL)&&(p2->dat=p1->dat))
{
p3=(pointer *)malloc(sizeof(struct pointer));
p3->dat=p1->dat;
p3->link=head3->link;
head3->link=p3;
}
p1=p1->link;
}
}

void cha(pointer *head1,pointer *head2, pointer *head3){ //计算集合1与集合2的差
pointer *p1,*p2,*p3;
p1=head1->link;
while(p1!=NULL)
{
p2=head2->link;
while((p2!=NULL)&&(p2->dat!=p1->dat))
p2=p2->link;
if(p2==NULL)
{
p3=(pointer *)malloc(sizeof(struct pointer));
p3->dat=p1->dat;
p3->link=head3->link;
head3->link=p3;
}
p1=p1->link;
}
}

main(){
pointer *head1,*head2,*head3;
head1=(pointer *)malloc(sizeof(struct pointer));
head1->link=NULL;
head2=(pointer *)malloc(sizeof(struct pointer));
head2->link=NULL;
head3=(pointer *)malloc(sizeof(struct pointer));
head3->link=NULL;
printf("输入集合1：\n");
readdata(head1);
printf("输入集合2：\n");
readdata(head2);
printf("集合1为：\n");
disp(head1);
printf("集合2为：\n");
disp(head2);
printf("集合1与集合2的并为:\n");
bing(head1,head2,head3);
disp(head3);
head3->link=NULL;
printf("集合1与集合2的交为:\n");
jiao(head1,head2,head3);
disp(head3);
head3->link=NULL;
printf("集合1与集合2的差为:\n");
cha(head1,head2,head3);
disp(head3);
}