c++ - 无法通过 C++ vector 进行解析

Question

我需要编写一个程序来输入由空格分隔的整数，例如:

4 4 5 8 8 9

程序然后获取这些数字并计算每个数字出现的次数，因此上述输入的输出将是:

The number 4 has 2 occurrence(s)
The number 5 has 1 occurrence(s)
The number 8 has 2 occurrence(s)
The number 9 has 1 occurrence(s)

我几乎已经弄清楚了，当我为数字没有用空格分隔的输入时它工作得很好(假设它们是 1 位整数，而不是我为最终版本所做的假设)但是一旦输入的数字之间有空格，它就不再有效。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <string>

using namespace std;

vector<int> parseString(string &s);
void parseVector(vector<int> &v);
int checkRepeats(vector<int> &v, int n);
void printVector(vector<int> &v);

int main()
{
vector<int> parsed;
vector<int> numbers;
string input;
bool keepGoing = true;
int nRepeats;                               // stores the number of times a number occurs, will constantly be overwritten

cout << "Enter some numbers: ";

while(true)
{   
    cin >> input;

    if(input == "stop" || input == "Stop")
    {
        break;
    }

    parsed = parseString(input);                                                    // parse input string to vector of ints
    parseVector(parsed);                                                            // send vector of ints to be checked for repeats
    //printVector(parsed);

    //cout << "\n";
}
}

void printVector(vector<int>&v)                                                     // not called right now, used for testing
{
    for(int i = 0; i < v.size(); i++)
    {
        cout << v.at(i) << " ";
    }
}

void parseVector(vector<int> &v)
{
    int x = 0;
int j = 0;
int nRepeats = 0;
int size = v.size();

for(int i = 0; i < size; i++)
{
    x = v.at(i);                                                                    // x equals the next element in vector 'v'
    nRepeats = checkRepeats(v, x);                                                  // count the number of times number 'x' occurs in vector 'v'
    //i += nRepeats - 1;
    cout << "The number: " << x << " has: " << nRepeats << " occurrence(s)\n";
}
}

int checkRepeats(vector<int> &v, int n)                                                 // counts the number of times a number is found in a given vector
{
    int nTimes = 0;
int size = v.size();

for(int i = 0; i < size; i++)
{

    if(v.at(i) == n)                                                                // match found, increment counter
    {
        nTimes++;
    }
}

return nTimes;
}

vector<int> parseString(string &s)
{
    vector<int> v;
int strLen = s.size();
int x;

for(int i = 0; i < strLen; i += 2)                                                  // increment by 2 to cut out white space from between the numbers
{
    x = s.at(i);
    x -= 48;                                                                        // subtract 48 from x, converts from ascii to int value
    v.push_back(x);
}

return v;
}

如果你转到该代码的第 88 行，并将循环计数器的增量从 i += 2 更改为 i++，它将完美地用于没有输入的情况空格，例如 445889 而不是 4 4 5 8 8 9

有人知道我可以尝试解决这个问题吗？

Answer 1

你可以尝试使用 map ( super 天真的版本:)

#include <map>
#include <iostream>

int main()
{
  int i;
  std::map<int, int> ints;

  while (std::cin >> i)
    ++ints[i];

  for (auto const& num : ints)
    std::cout <<
      "The number " << 
      num.first <<  
      " has " <<  
      num.second << 
      " occurrence(s)\n";
}