STL finish 集合交、差、并、补运算

分析

集合的运算嘛，使用链表啊；YOU are right. 不过，最近学了STL模板库，可不可以用模板做呢？YES.而且很容易想到STL set.

why

set模板拥有强大的自动排序功能，减化排序与去重操作

think，不需要你编写冗杂的排序与去重，是不是可以节约很多时间呢？

节省了时间，赶紧去霸占篮球场啊（嘿嘿，我是詹皇的球迷哦）

Follow is my code in details

#include <iostream>
#include <cstdlib>
#include <windows.h>
#include <set>
using namespace std;

set<char> Collection_Quan;
set<char> Collection_A;
set<char> Collection_B;

void Menu();
void Set_Q();
void Set_A();
void Set_B();
void Set_Jihe();
void Bingji();//并集
void Jiaoji();//交集
template <class T>
void Chaji(const T&, const T&); //差集

template <class T>
void show(const T& Input)//T是迭代器
{
for(auto im = Input.cbegin();im != Input.cend();im++)
cout << *im << " ";
cout << endl;
}

int main()
{
system("color 0E");
cout << "程序功能求集合A与B的交集、并集、差集、补集" << endl;

Set_Jihe();
Menu();
//Bingji();
//Jiaoji();
//Chaji();
system("pause");
return 0;
}

void Bingji()//求A 与 B 并集
{
set <char> bingji;
bingji.insert(Collection_A.begin(), Collection_A.end());
bingji.insert(Collection_B.begin(), Collection_B.end());
cout << "A 并 B = :    ";
show(bingji);
}

void Jiaoji()//求A 与 B 交集
{
set<char> jiaoji;
for(auto iElem = Collection_A.begin();iElem != Collection_A.end();iElem++)
for(auto jElem = Collection_B.begin();jElem != Collection_B.end();jElem++)
if(*iElem == *jElem)
{
jiaoji.insert(*iElem);
break;
}
cout << "A 交 B = :    ";
show(jiaoji);
}

template <class T>
void Chaji(const T& First_Co, const T& Second_Co)//求A 与 B 差集
{
set<char> chaji;
for(auto iElem = First_Co.cbegin();iElem != First_Co.cend();iElem++)
{
bool flag = true;
for(auto jElem = Second_Co.cbegin();jElem != Second_Co.cend();jElem++)
if(*iElem == *jElem)//找出A中存在， B中不存在的元素
{
flag = false;
break;
}
if(flag) chaji.insert(*iElem);
}
if(First_Co == Collection_A) cout << "A -  B = :    ";
else cout << "B -  A = :    ";
show(chaji);
}

void Menu()
{
cout << "***求集合A与B的交，并、补集***" << endl;
cout << "1.交集  2.并集   3.A - B   4.B- A  5.退出" << endl;
cout << "******************************" << endl;
cout << "choose :" << endl;
int n;
while(1)
{
cin >> n;
switch(n)
{
case 1: Jiaoji();break;
case 2: Bingji();break;
case 3: Chaji(Collection_A, Collection_B); break;//A - B
case 4: Chaji(Collection_B, Collection_A); break;//B - A
case 5: exit(1);
default:break;
}
}

}
void Set_Jihe()
{
Set_Q();
Set_A();
Set_B();

}
void Set_Q()
{
char Elem_x;
int Q_count;
auto cheak = Q_count;
cout << "请输入全集Q包含元素的个数：";
cin >> cheak;
while(sizeof(cheak) != sizeof(int))
{
cout << "输入错误，请重新输入： " << endl ;//此处要有错误处理，判断
cin >> cheak;
}
Q_count = cheak;
cout << "请输入全集Q包含元素值：    ";
while(Q_count--)
{
cin >> Elem_x;
Collection_Quan.insert(Elem_x);
}
cout << endl;
}

void Set_A()
{
char Elem_x;
int A_count;
cout << "请输入集合A包含元素的个数：";
cin >> A_count;
cout << "请输入集合A包含元素值：    ";
while(A_count--)
{
cin >> Elem_x;
Collection_A.insert(Elem_x);
}
cout << endl;
}

void Set_B()
{
char Elem_x;
int B_count;
cout << "请输入集合B包含元素的个数：";
cin >> B_count;
cout << "请输入集合B的元素值：      ";
while(B_count--)
{
cin >> Elem_x;
Collection_B.insert(Elem_x);
}
cout << endl;
}
//if the code is terrible, I hope you can tell me. :)

Be quick to finish your code amzingly.



HERE 詹皇 ：）



哈哈，当然是要写完成code tasks nuo.加油.