在C++中寻找解决此问题的特定设计模式

我在以下一点更新您的代码：-

1. 将类转换为单一实例，以便数据为每种类型维护单个映射
2. 将地图更改为unorder_map

#define INPUT 8 
using namespace std;
/*use class to store single slip data*/
class Note {
public:
string Tital;
string Text;
vector<string> vec;
Note(){ //constructor to initialize data zero
Tital = "";
Text = "";
}
};
/*create a singalton pattern class so that create only one data storage*/
class storyBoard{
public:
static storyBoard* getInstance(){
storyBoard* Instance= instance.load();
if ( !Instance ){
std::lock_guard<std::mutex> myLock(lock);
Instance= instance.load();
if( !Instance ){
Instance= new storyBoard();
instance.store(Instance);
}
}   
return Instance;
}
void AddNote(string Tital,string Text,vector<string> vec );
void RemoveByTital(string &tital);
void PrintStoredData();
Note* searchByTitle(string titleSearch);
Note* searchByText(string text_);
vector<Note*> searchByTag(string titleSearch);
void printSlip(Note *tm);
private:
storyBoard()= default;
~storyBoard()= default;
storyBoard(const storyBoard&)= delete;
storyBoard& operator=(const storyBoard&)= delete;
static std::atomic<storyBoard*> instance;
static std::mutex lock;
std::unordered_map<string,Note *> TitalMap;
std::unordered_map<string,Note *> TextMap;
std::unordered_map<string,std::vector<Note *> > TagsMap;
};

std::atomic<storyBoard*> storyBoard::instance;
std::mutex storyBoard::lock;
Note* storyBoard::searchByTitle(string titleSearch){
auto it_v = TitalMap.find(titleSearch);
if (it_v != TitalMap.end()){
cout<< "Tital search result is below:-"<<endl;
return it_v->second;
} else {
cout <<"data "<<titleSearch << "  Not found"<<endl;
return NULL;
}
}
Note* storyBoard::searchByText(string titleSearch){
auto it_v = TextMap.find(titleSearch);
if (it_v != TextMap.end()){
cout<< "Text search result is below:-"<<endl;
return it_v->second;
} else {
cout <<"data "<<titleSearch << "  Not found"<<endl;
return NULL;
}
}
vector<Note*> storyBoard::searchByTag(string tagSearch){
auto it_v = TagsMap.find(tagSearch);
if (it_v != TagsMap.end()){
cout<< "Tag search result is below:-"<<endl;
return it_v->second;
} else {
cout <<"data "<<tagSearch << "  Not found"<<endl;
vector<Note*> del;
return del;
}
}

void storyBoard::AddNote(string Tital, string Text, vector<string> v){
Note *note = new Note;
note->Tital = Tital;
note->Text = Text;
note->vec = v;
TitalMap[note->Tital] = note;
TextMap[note->Text] = note;
for (auto it = note->vec.begin(); it != note->vec.end(); ++it){
//check that is tags already 
auto it_v = TagsMap.find(*it);
if (it_v != TagsMap.end()){
it_v->second. push_back(note);
} else {
vector<Note *> &v = TagsMap[*it];
v.push_back(note);
}
}   
}
void storyBoard::printSlip(Note *tm){
cout << "Tital=" << tm->Tital <<endl 
<< "Text=" <<  tm->Text <<endl
<< "Tags = ";
for (auto it = tm->vec.begin(); it != tm->vec.end(); ++it){
cout<< *it<<"t";
}    
cout<<endl<<endl;
}
void storyBoard::PrintStoredData(){
for(auto tm : TitalMap){
printSlip(tm.second);
}
cout<<endl; 
}
/**temporary function only use for testing*/
void TestCase(){    
storyBoard *Sb = storyBoard::getInstance();
Note* obj = Sb->searchByTitle("Tital-3");
if(obj != NULL){
Sb->printSlip(obj);
}
obj = Sb->searchByText("Text-4");
if(obj != NULL){
Sb->printSlip(obj);
}
vector<Note *> vec = Sb->searchByTag("tag-3");
if(vec.size() !=0){
for (auto it = vec.begin(); it != vec.end(); ++it){
//cout<<(*it)->Tital << "t";
Sb->printSlip(*it);
} 
}   
}
/**temporary function only use for testing*/
void feed_data_for_testing(){
storyBoard *Sb = storyBoard::getInstance();
vector<string> tags ;
int count =INPUT;
for(int i =1;i<=count;i++){
string tital = "Tital-" + std::to_string(i);
string text = "Text-" + std::to_string(i);
tags.clear();
for(int j =1;j<=i;j++){
string tag_ = "tag-" +  std::to_string(j);
tags.push_back(tag_);
}
Sb->AddNote(tital,text,tags);
}
}
int main() {
storyBoard *Sb = storyBoard::getInstance();
feed_data_for_testing();
Sb->PrintStoredData(); /*Print all contain data */
cout<<"************* From Here start searching ************"<<endl;
TestCase();
return 0;
}

我认为您的面试官错误地使用了术语设计模式来代替术语习语。

代码的一个主要问题(可能是拒绝的原因)是使用经典 c++ 习惯用法的内存处理：

• 使用智能指针管理笔记的生命周期
• 处理StoryBoard对象的复制/分配(/移动)的 3 法则(或 5)

请注意，使用智能指针只是管理内存的一种方法。您也可以使用其他习语：

• 竞技场和引用来自哈希地图中竞技场的笔记
• 。

解决此问题后，您会遇到一些小问题：

• 返回指针而不是引用，StoryBoard是内存的所有者，不应返回调用者可能无意中释放的指针。
• 没有常量访问器(返回常量引用)
• 可分解的重复代码

如果我在解释面试官所说的话时没有记错的话，这个问题应该移到 codereview.stackexhange.com