Question

我有一个这样的数字/字符串（我不知道如何将int转换为字符串和从字符串转换）

000000122310200000223340000700012220000011411000000011011271043334010220001127100003333201000001000070005222500233400000000000000000000

我需要做的是将0之间的数字分开，所以我得到像

这样的字符串

“12231” “22334” “7” “1222”

依此类推，然后我需要将它们转换为int。（基本上我已经搜索过如何进行转换无效）

有人可以帮忙吗？

谢谢！

Answer 1

std::getline采用可选的分隔符。通常情况下，这将是换行符（因此是getline），但您可以使用0。

Answer 2

解决方案std::getline读数为＆＃39; 0＆＃39;

live

// First create a string stream with you input data
std::stringstream ss("000000122310200000223340000700012220000011411000000011011271043334010220001127100003333201000001000070005222500233400000000000000000000");;

// Then use getline, with the third argument it will read untill zero character
// is found. By default it reads until new line.
std::string line;
while(std::getline(ss, line, '0')) {

    // In case there are no data, two zeros one by one, skip this loop
    if ( line.empty() )
       continue;

    // now parse found data to integer
    // Throws excepions if bad data, consult: http://en.cppreference.com/w/cpp/string/basic_string/stol
    int n = std::stoi(line);
   std::cout << n << "\n";
}

Answer 3

如果C ++ 11对你有好处，那么这应该可行

#include <vector>
#include <string>
#include <sstream>

std::vector<int> split(const std::string& s, char delim)
{
    std::vector<int> res;
    std::stringstream ss(s);
    std::string sub;
    while (std::getline(ss, sub, delim)) 
    {
        int val = std::stoi(sub);  // c++11
        res.push_back(val);
    }
    return res;
}

Answer 4

只需从字符串中进行迭代，然后检查它是否＆＃39; 0＆＃39; 0＆＃39; 0或不。

#include<bits/stdc++.h>
using namespace std;

int main() {
   char a[] = "000000122310200000223340000700012220000011411000000011011271043334010220001127100003333201000001000070005222500233400000000000000000000";
   vector<int> nums;

   int tmp = 0;
   for(int i=0;a[i];i++) {
       if(a[i] != '0') {
          tmp *= 10;
          tmp += a[i]-'0';
       } else {
          if(tmp != 0){
              nums.push_back(tmp);
              tmp = 0;
          }
       }
   }

   for(int i=0;i<nums.size();i++)
      cout << nums[i] << endl;
}

Answer 5

这是一个可能的解决方案：

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

std::vector<int> split(const std::string& x, char separator)
{
    std::stringstream stream(x);
    std::vector<int> numbers;
    std::string line;
    while(std::getline(stream, line, separator)) {
        if (!line.empty()){
            int n = std::stoi(line);
            numbers.push_back(n);
        }
    }
    return numbers;
}

int main()
{
    std::string myStringValue("000000122310200000223340000700012220000011411000000011011271043334010220001127100003333201000001000070005222500233400000000000000000000");
    std::vector<int> values = split(myStringValue,'0');
    for(auto i : values) {
        std::cout << i << std::endl;
    }
}

由0分隔的C ++字符串解析

5 个答案: