在C ++ 98中优化向量的过滤

Question

I was given a following task: create a text file which contains men, women and unknown names. Try to implement a filter that will compare those names with real names (so 3 files on input in total: men, women and let's say renters). While filtering put matching names in their appropriate containers. It seemed to me pretty straight forward so I did it the way I provide underneath.

My question is: Is there a way how to optimize this code?

I tried to use abstract classes and create 4 different objects (Man, Woman, Known, Unknown) based on abstract entity. But the amount of code was still large for such a simple task. Another idea was to use lambda expressions, but I am limited to C++ 98 only.

I think I am overthinking it...

#include <fstream>
#include <iostream>
#include <vector>
#include <string>

int main()
{
    std::ifstream men("resources/men_names.txt");
    std::ifstream women("resources/women_names.txt");
    std::ifstream renters("resources/renter_names.txt");

    std::vector<std::string> menNames;
    std::vector<std::string> womenNames;
    std::vector<std::string> renterNames;
    std::vector<std::string> knownRenters;
    std::vector<std::string> unknownRenters;

    std::string name;

    while (men >> name)
        menNames.push_back(name);

    men.close();

    while (women >> name)
        womenNames.push_back(name);

    women.close();

    while (renters >> name)
        renterNames.push_back(name);

    renters.close();

    std::vector<std::string>::iterator itMen;
    std::vector<std::string>::iterator itWomen;
    std::vector<std::string>::iterator itRenters;

    for (itRenters = renterNames.begin(); itRenters != renterNames.end(); itRenters++)
    {
        bool found = false;

        for (itMen = menNames.begin(); itMen != menNames.end(); itMen++)
        {
            if ((*itMen) == (*itRenters))
            {
                found = true;
                knownRenters.push_back((*itMen));
            }
        }
        if (!found)
        {
            for (itWomen = womenNames.begin(); itWomen != womenNames.end(); itWomen++)
            {   
                if ((*itWomen) == (*itRenters))
                {
                    found = true;
                    knownRenters.push_back((*itWomen));
                }
            }   
        }
        if (!found)
            unknownRenters.push_back((*itRenters));
    }

    std::cout << knownRenters.size() << '\n';
    std::cout << unknownRenters.size() << '\n';

    std::cin.get();

    return 0;
}

Answer 1

A shortening of your existing code. This should be all C++98

#include <fstream>
#include <iostream>
#include <vector>
#include <string>
#include <set>
#include <iterator>

int main()
{
    std::ifstream men("resources/men_names.txt");
    std::ifstream women("resources/women_names.txt");
    std::set<std::string> peopleNames;

    peopleNames.insert(std::istream_iterator<std::string>(men), std::istream_iterator<std::string>());
    peopleNames.insert(std::istream_iterator<std::string>(women), std::istream_iterator<std::string>());

    std::ifstream renters("resources/renter_names.txt");
    std::vector<std::string> knownRenters;
    std::vector<std::string> unknownRenters;

    for (std::string name; renters >> name; )
    {
        if (peopleNames.count(name))
            knownRenters.push_back(name);
        else
            unknownRenters.push_back(name);
    }

    std::cout << knownRenters.size() << '\n';
    std::cout << unknownRenters.size() << '\n';

    std::cin.get();

    return 0;
}

Answer 2

You don't care whether a renter is male or female, just if they have a recognised name. So, don't store two flat vectors of known names, store a single std::set<std::string> of all recognised names (or a sorted vector, or a std::unordered_set if you're ever allowed to use C++11).

Then, instead of doing (at most) two linear searches per renter, you can do a single logarithmic-time lookup (or constant-time for the C++11 version).

You also don't seem to care about the names of the renters who were recognised (or not), so don't keep two result vectors: just increment a known or unknown counter.