Question

注意：我无法在算法库

中使用地图或任何内容

我只有主要功能，但我完全迷失了我应该写的功能

#include <string>
#include <iostream>


using namespace std;


int* count(const string& s);

int main() {

    string userinput = "random word";

    int *counts = count(userinput);

    for (int i = 0; i < 11; i++) {
        cout << "Letter " << i << " occured " << counts[i] << " times.";
    }




    system("pause");

    return 0;
}



int* count(const string& s) {

    for (int i = 0; i < 11; i++)
    {
        return s[i];
    }
}

int * count函数不正确，无法编译。如何编写一个可以返回出现的函数？

Answer 1

如何使用int *计算字符串中每个数字的出现次数在c ++中count（const string＆amp; s）？注意：我无法使用地图或任何内容算法库

有几点需要注意：

总是不好主意，如果我们不采用最简单的方法解决问题，那么老师是这种情况的推动者会感到很难过。
不需要int*。它不以任何方式提供您的解决方案。如果它是数组来计算字符串的字符，那么它可能是有意义的（即指向数组的指针）。
尽量避免使用using namespace std;

如果您不允许使用std::map<char, int>（提示替代解决方案），您可以执行以下操作。

我对此有所评论，以便更好地理解。

#include <string>
#include <iostream>
#include <cstddef>

// consider 128 ASCII decimal and their coresponding character codes
int charASCIIArray[128] = {0};

void count_char(const std::string& s)
{
   for(const auto& it: s)
   {
        if(('A' <= it && it <= 'Z') ||     // if char between (A,B,....,Z) or
           ('a' <= it && it <= 'z') )      //         between (a,b,....,z)
           charASCIIArray[static_cast<int>(it)]++; // we count each corresponding array then
   }
}

int main()
{
   std::string userinput = "random words WITH *- aLl";

   count_char(userinput);
   for(std::size_t index = 0; index < 128; ++index)
      if(charASCIIArray[index] != 0)
        std::cout << "Letter " << static_cast<char>(index)  // convert back to char
                  << " occured " << charASCIIArray[index] << " times.\n";

   return 0;
}

在此处查看实时操作：https://www.ideone.com/uFG2HJ

Letter H occured 1 times.
Letter I occured 1 times.
Letter L occured 1 times.
Letter T occured 1 times.
Letter W occured 1 times.
Letter a occured 2 times.
Letter d occured 2 times.
Letter l occured 1 times.
Letter m occured 1 times.
Letter n occured 1 times.
Letter o occured 2 times.
Letter r occured 2 times.
Letter s occured 1 times.
Letter w occured 1 times.

更新

受到@Fei Xiang评论的启发，这里有两个过度设计的解决方案：

首先，返回指向动态数组的指针（这将符合实际值问题要求）：https://www.ideone.com/ouHqK4

#include <string>
#include <iostream>
#include <cstddef>

int* count_char(const std::string& s)
{
    // consider 128 ASCII decimal and their coresponding character codes
    int *charASCIIArray = new int[128]{0};
    for(const auto& it: s)
    {
        if(('A' <= it && it <= 'Z') ||     // if char between (A,B,....,Z) or
           ('a' <= it && it <= 'z') )      //         between (a,b,....,z)
           charASCIIArray[static_cast<int>(it)]++; // we count each corresponding array then
    }
    return charASCIIArray;
}

int main()
{
    std::string userinput = "random words WITH *- aLl";

    int *charASCIIArray = count_char(userinput);
    for(std::size_t index = 0; index < 128; ++index)
        if(charASCIIArray[index] != 0)
        std::cout << "Letter " << static_cast<char>(index)  // convert back to char
                  << " occured " << charASCIIArray[index] << " times.\n";

    delete[] charASCIIArray;
    return 0;
}

其次使用智能指针（std::unique_ptr）：https://www.ideone.com/dfc62J

#include <string>
#include <iostream>
#include <cstddef>
#include <memory>
#include <utility>

std::unique_ptr<int[]> count_char(const std::string& s)
{
    // consider 128 ASCII decimal and their coresponding character codes
    std::unique_ptr<int[]> charASCIIArray = std::unique_ptr<int[]>(new int[128]{0});
    for (const auto& it : s)
    {
        if (('A' <= it && it <= 'Z') ||     // if char between (A,B,....,Z) or
            ('a' <= it && it <= 'z'))       //         between (a,b,....,z)
            charASCIIArray[static_cast<int>(it)]++; // we count each corresponding array then
    }
    return std::move(charASCIIArray);
}

int main()
{
    std::string userinput = "random words WITH *- aLl";

    std::unique_ptr<int[]> charASCIIArray = count_char(userinput);
    for (std::size_t index = 0; index < 128; ++index)
        if (charASCIIArray[index] != 0)
            std::cout << "Letter " << static_cast<char>(index)  // convert back to char
            << " occured " << charASCIIArray[index] << " times.\n";

    return 0;
}

Answer 2

不太确定你想要这个答案，但我希望它有所帮助你可以把它分成单独的功能，但我一次性完成了

-0.028450677

如何使用c ++中的int * count（const string＆amp; s）来计算字符串中每个数字的出现次数？

2 个答案: