如何将带有尾随 x 的数字字符串转换为无符号数字列表

Question

我需要将 4 个字符的字符串转换为无符号数字列表例如 222X = [2220, 2221, 2222, 2223, 2224, 2225, 2226, 2227, 2228, 2229 ] 类似地，我应该能够转换 22XX（100 个数字）、2XXX（应该产生 1000 个）数字。有什么快速的方法吗

我有以下解决方案，但不是很干净..

std::list<unsigned> stringToCode(std::string fn)
{
   std::list<unsigned> codes;
   unsigned count = std::count(fn.begin(), fn.end(), 'X');

   unsigned none_x = std::stoi(fn);
   unsigned numbers_to_generate = std::pow(10, count);

   unsigned overrule = none_x * numbers_to_generate;
   for (int i = 0; i < numbers_to_generate; i++) {
      unsigned fnumber = none_x * std::pow(10, count) + i;

      codes.push_back(fnumber);
   }

   return codes;
  }

int main() 
{

  std::string number = "4XXX";

  std::list<unsigned> codes = stringToCode(number);

   for (const auto code : codes) {
      std::cout << code << std::endl;
   }

   return 0;
}

Answer 1

创建两个变量：

std::string maxVal = fn; 
std::replace(maxVal, 'X', '9'); 
std::string minVal = fn;
std::replace(minVal, 'X', '0');

现在你可以循环使用

for (auto i = std::stoi(minVal), j = std::stoi(maxVal); i <= j; ++i) {
    codes.push_back(i);
}

整个代码

#include <algorithm>
#include <iostream>
#include <list>

std::list<unsigned> stringToCode(std::string fn) {
   std::string maxVal = fn; 
   std::replace(std::begin(maxVal), std::end(maxVal), 'X', '9'); 
   std::string minVal = fn;
   std::replace(std::begin(minVal), std::end(minVal), 'X', '0');
   std::list<unsigned> codes;
   for (auto i = std::stoi(minVal), j = std::stoi(maxVal); i <= j; ++i) {
       codes.push_back(i);
   }

   return codes;
}

int main() {
  std::string number = "4XXX";

  std::list<unsigned> codes = stringToCode(number);

   for (const auto code : codes) {
      std::cout << code << std::endl;
   }

   return 0;
}

如何将带有尾随 x 的数字字符串转换为无符号数字列表

how to convert a string of number with trailing x's into a list of unsigned numbers

c++

arrays

string

int