如何用链表实现集合的交集，并集，差集
\u7528\u94fe\u8868\u5b9e\u73b0\u96c6\u5408\u7684\u4ea4\u96c6\u5e76\u96c6\u5dee\u96c6#include
#include
#include
#include
using namespace std;
int main()
{
int a[]={1,5,8,12,5,-5,32};
int b[]={3,5,1,-3,10};
list set1(a,a+sizeof(a)/sizeof(int));
list set2(b,b+sizeof(b)/sizeof(int));
list result;
set1.sort();
set2.sort();
//\u4ea4\u96c6
set_intersection(set1.begin(),set1.end(),set2.begin(),set2.end(),back_inserter(result));
copy(result.begin(),result.end(),ostream_iterator(cout," "));
cout<<endl;
result.clear();
//\u5e76\u96c6
set_union(set1.begin(),set1.end(),set2.begin(),set2.end(),back_inserter(result));
copy(result.begin(),result.end(),ostream_iterator(cout," "));
cout<<endl;
result.clear();
//\u5dee\u96c6
set_difference(set1.begin(),set1.end(),set2.begin(),set2.end(),back_inserter(result));
copy(result.begin(),result.end(),ostream_iterator(cout," "));
return 0;
}
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#define maxsize 100
struct Sqlist
{
int elem[maxsize];
int length;
};
void difference(Sqlist &la,Sqlist lb)
{ //求A与B的并集
int i,j,e;
for(i=0;i<lb.length;i++)
{
e=lb.elem[i]; j=0;
while((j<la.length)&&(la.elem[j]!=e)) j++;
if(j==la.length){
la.elem[la.length]=e;la.length++;}
}
}
//求A与B的交集
void Intersection(Sqlist &la,Sqlist lb)
{int i,j;
static int n=0;
for(i=0;i<lb.length;i++)
{ j=0;
while(j<la.length&&la.elem[j]!=lb.elem[i])
j++;
if(la.elem[j]==lb.elem[i])
{
la.elem[n]=lb.elem[i];
n++;
}
}
void main()
{
int i;
struct Sqlist la,lb,lc;
printf("请输入集合A、集合B的元素个数\n");
scanf("%d %d",&la.length,&lb.length);
printf("请输入集合A的元素\n");
for(i=0;i<la.length;i++)
scanf("%d",&la.elem[i]);
printf("请输入集合B的元素\n");
for(i=0;i<lb.length;i++)
scanf("%d",&lb.elem[i]);
difference(la,lb);
printf("并集中的元素为：\n");
for(i=0;i<la.length;i++)
printf("%-4d",la.elem[i]);
printf("\n");
Intersection(la,lb);
printf("交集中的元素为：\n");
for(i=0;i<n;i++)
printf("%-4d",la.elem[i]);
printf("\n");
}
我只会顺序表的
#include <algorithm>
#include <iostream>
#include <iterator>
#include <list>
using namespace std;
int main()
{
int a[]={1,5,8,12,5,-5,32};
int b[]={3,5,1,-3,10};
list< int > set1(a,a+sizeof(a)/sizeof(int));
list< int > set2(b,b+sizeof(b)/sizeof(int));
list< int > result;
set1.sort();
set2.sort();
//交集
set_intersection(set1.begin(),set1.end(),set2.begin(),set2.end(),back_inserter(result));
copy(result.begin(),result.end(),ostream_iterator< int >(cout," "));
cout<<endl;
result.clear();
//并集
set_union(set1.begin(),set1.end(),set2.begin(),set2.end(),back_inserter(result));
copy(result.begin(),result.end(),ostream_iterator< int >(cout," "));
cout<<endl;
result.clear();
//差集
set_difference(set1.begin(),set1.end(),set2.begin(),set2.end(),back_inserter(result));
copy(result.begin(),result.end(),ostream_iterator< int >(cout," "));
return 0;
}
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