Question

C ++编程任务

所以我有一个字符串列表

static const string mCodes[] = {
    "AAD - Architecture Design",
    "AAE - Architecture",
    "AAI - Interior Design",
    "AAL - Landscape Architecture",
    "AAP - Urban Planning",
    "AAS - Afro-American Studies",
    "ABS - Architecture Building Science",
    "ACA - Architecture Construction Mgmt",
    "ACC - Accounting",
    "AES - Aerospace Studies"};

我让用户使用学生的专业代码（例如，studentMajor = ACC）将studentMajor定义为2-4个字符串

我需要搜索字符串列表，如果它在该字符串列表中找到ACC，则会将studentMajor设置为在其中找到ACC的完整字符串。

if( student variable is found ) then
    studentVariable = "ACC - Accounting"
else
    error, improper student major entered

我假设我需要使用std :: find，但是我不确定如何在该列表中查找“ ACC”，然后将完整的字符串“ ACC”附加并存储到变量中

我认为对std :: string :: find和std :: string :: substr的操作很简单，但是我不确定如何解决这个问题。

谢谢

Answer 1

一种方法是使用std::for_each并仅在每个完整字符串中搜索提供的字符串。我还将使用std::vector中的std::string来存储字符串。示例：

#include <iostream>
#include <algorithm>
#include <vector>

const std::vector<std::string> mCodes = {
    "AAD - Architecture Design",
    "AAE - Architecture",
    "AAI - Interior Design",
    "AAL - Landscape Architecture",
    "AAP - Urban Planning",
    "AAS - Afro-American Studies",
    "ABS - Architecture Building Science",
    "ACA - Architecture Construction Mgmt",
    "ACC - Accounting",
    "AES - Aerospace Studies"};

std::vector<std::string> find_matches(const std::string& in) {
    std::vector<std::string> rv;

    // loop through all mCodes
    std::for_each(mCodes.begin(), mCodes.end(),
        // call lambda function for each mCode
        [&](const std::string& full) {
            // check if the "in" string can be found in "full"
            if(full.find(in)!=std::string::npos)
                // if so, save "full" to "rv"
                rv.push_back(full);;
        }
    );

    // return a vector of all found strings
    return rv;
}

如果要使用2-4个字母代码的精确匹配，则可以使用std::unordered_map，在代码和课程全名之间进行映射。示例：

#include <iostream>
#include <vector>
#include <unordered_map>

const std::unordered_map<std::string, std::string> mCodes = {
    {"AAD", "Architecture Design"},
    {"AAE", "Architecture"},
    {"AAI", "Interior Design"},
    {"AAL", "Landscape Architecture"},
    {"AAP", "Urban Planning"},
    {"AAS", "Afro-American Studies"},
    {"ABS", "Architecture Building Science"},
    {"ACA", "Architecture Construction Mgmt"},
    {"ACC", "Accounting"},
    {"AES", "Aerospace Studies"}
};

int main(int argc, char* argv[]) {
    std::vector<std::string> args(argv+1, argv+argc);

    for(auto& str : args) {
        auto it = mCodes.find(str);
        if(it != mCodes.end()) {
            std::cout << it->first << " - " << it->second << "\n";
        }
    }
}

Answer 2

我通过在for循环中运行它解决了这个问题。

for(int i = 0; i < 221; i++){
    if(mCodes[i].find(testMajor) == string::npos){
        testMajor = mCodes[i];
        return testMajor;
    }
}

其中221是数组mCode的完整大小。这只是手动搜索mCodes的每个元素，如果搜索返回正数，则将testMajor设置为数组的该元素：）

谢谢你们

Answer 3

您的问题将使自己适应地图，该地图将课程代码用作地图的关键。这是一个简单的例子。

#include <iostream>
#include <string>
#include <map>

static const std::string mCodes[] = {
    "AAD - Architecture Design",
    "AAE - Architecture",
    "AAI - Interior Design",
    "AAL - Landscape Architecture",
    "AAP - Urban Planning",
    "AAS - Afro-American Studies",
    "ABS - Architecture Building Science",
    "ACA - Architecture Construction Mgmt",
    "ACC - Accounting",
    "AES - Aerospace Studies"};

int main()
{
    std::map<std::string, std::string> course_map;

    for(auto const& course: mCodes) {
        std::string code = course.substr(0, course.find(" "));
        course_map.insert(std::pair<std::string, std::string>(code, course));
    }

    if(course_map.find("ACC") != course_map.end() ) {
        std::cout << "course is " << course_map["ACC"] << std::endl;
    } else {
        std::cout << "ACC not found." << std::endl;
    }
    return 0;
}

在字符串列表中查找字符串，然后将字符串设置为找到的字符串？

3 个答案: