一、概述
案例:c++ stl之set集合练习
二、代码示例
#include <iostream> #include <set> #include <string> using namespace std; //打印set元素 void printSet(set<int> &s){ for(set<int>::iterator it=s.begin();it!=s.end();it++){ cout << *it<<" "; } cout <<endl; } void test(){ set<int> s; //向set集合中添加元素 s.insert(10); s.insert(20); s.insert(30); s.insert(40); s.insert(30); printSet(s); if(s.empty()){ cout << "set is Null"<<endl; }else{ cout << "set is Not Null"<<endl; } //删除元素为30的元素 s.erase(30); printSet(s); } void test2(){ set<int> s; s.insert(10); s.insert(20); s.insert(30); s.insert(40); s.insert(50); s.insert(60); set<int>::iterator pos = s.find(30); if(pos!=s.end()){ cout << "I found element:"<<*pos<<endl; }else{ cout <<"I Not Found Element"<<endl; } int num = s.count(40); cout <<"key=40 size is :"<<num<<endl; set<int>::iterator pos2 = s.lower_bound(30); if (pos2 != s.end()) { cout << "lower_bound value is:" << *pos2 << endl; } else { cout << "lower_bound value not fond" << endl; } pos2 = s.upper_bound(30); if (pos2 != s.end()) { cout << "upper_bound value is :" << *pos2 << endl; } else { cout << "upper_bound value not found" << endl; } // pair<set<int>::iterator,set<int>::iterator> ret = s.equal_range(30); if(ret.first!=s.end()){ cout << "equal_range lower_bound value is:" << *ret.first << endl; }else { cout << "equal_range lower_bound value not found" << endl; } if (ret.second != s.end()) { cout << "equal_range upper_bound value is:" << *ret.second << endl; } else { cout << "equal_range upper_bound value not found" << endl; } } void test3(){ pair<string,int> p("tony",30); cout << "name: "<<p.first<< " age:"<<p.second<<endl; pair<string,int> p2 = make_pair("luoluoyang",3); cout << "name: "<< p2.first<<" age: "<<p2.second<<endl; } void test4(){ set<int> s; pair<set<int>::iterator,bool> ret = s.insert(10); if(ret.second){ cout << "insert success"<<endl; }else{ cout << "insert fail"<<endl; } ret = s.insert(10); if(ret.second){ cout <<"second insert success"<<endl; }else{ cout <<"second insert fail"<<endl; } printSet(s); } void test5(){ } /** * * set集合的特性:可以过滤掉重复的元素 * */ int main(int argc, char const *argv[]) { // test(); // test2(); // test3(); test4(); return 0; }