Please help me with two issues I have on a program.

[korno917]

...

[Duthomhas]

[edit]

It seems that korno917 has entitlement issues on a public forum, so good riddance.

His homework was to take a string of alphabetic letters and do some run-length encoding. For example: “aaabcc” → “3ab2c”.

Here’s a recursive version using ONLY STRINGS, LOL:

#include <iostream>
#include <sstream>
#include <string>

std::string compress( const std::string& s )
{
  if (s.empty()) return "";
  auto n = s.find_first_not_of( s[0] );
  if (n == s.npos) n = s.size();
  if (n > 1) return std::to_string( n ) + s[0] + compress( s.substr( n ) );
  else       return                       s[0] + compress( s.substr( n ) );
}

int main()
{
  std::string data;
  while (std::cin >> data)
  {
    std::cout << "The compressed data is: " << compress( data ) << "\n";
  }
}

Turn that in and see if your prof fails you for academic dishonesty.

My original response follows.
[/edit]

You have named your data string “runLength”. It is not the run length. It is the data. JSYK.

If you are using cin >> data then you are automatically breaking on spaces! So all you need is an outer loop to print your message.

while (cin >> data)  // as long as there is space-separated input...
{
  std::cout << "The compressed data is: " << compress( data ) << "\n";
}

BTW, no need for vectors. A string is nicer to use than a vector<char>.

I do not understand why you need a separate string with only “aea” in it. This does not represent the data in any way. Have I misunderstood a need?


BTW, it is not cheating to ask for help.
It is cheating to have someone do your work for you.

Hope this helps.

[Duthomhas]

(1)
That's fine. Use vectors.

Create two of them: one for the compressed data, one for the letters. Every time you add a letter to the compressed data, also add it to the other. Every time you add a number, don't.

To add to the vectors, instead of using cout, just push_back().

std::vector <char> compressed_data;
std::vector <char> letters;

for (...)
{
  ...
  compressed_data.push_back( number_of_times );
  compressed_data.push_back( letter );
  letters.push_back( letter );
  ...
}

(2)
Correct.

[Duthomhas]

(1)
Ah, just remember the last, and count how many duplicates.

// pseudocode

while (characters left)
{
  int count = 1;
  char last = current character;
  current++;
  while (current character == last)
  {
    count++;
    current++;
  }

  count → string.
  add each character of count string to compressed_data vector
  add last to compressed_data vector and letters vector
}

Is this not more-or-less what you had?

(2)
It works just fine. I gave you a working example — that assumed strings. Here is one assuming vectors:

while (cin >> data)
{
  vector<char> compressed_data, letters;
  compress( data, compressed_data, letters );
  cout << "The compressed data is: ";
  for (auto c : compressed_data)
    cout << c;
  cout << "\n";
}

This assumes a function that looks like:

void compress( string data, vector<char>& compressed_data, vector<char>& letters )
{
  ...
}

[zapshe]

You could use a nested loop:

#include <iostream>
#include <string>
#include <vector>

using namespace std;

int main()
{
	string runLength;

	vector<int> lengths;
	vector<char> chars;

	cout << "Enter the data to be compressed: ";
	cin >> runLength;

	//Do Whatever You Need For WhiteSpace


	//"i" is to parse the string, "t" is keeping track of loop count
	for (int i = 0, t = 0; i < runLength.size(); i++, t++)
	{
		int track = 0; //See how many letters to skip in string

		lengths.push_back(1); //Every letter at least appears once
		chars.push_back(runLength[i]); //Add each letter

		for (int j = i + 1; j < runLength.size(); j++)
		{
			//BREAK if the next letter doesn't match
			if (runLength[i] != runLength[j]) 
				break;

			//If it does match, add to the length
			lengths[t]++;
			//And also keep track of how many times this happens
			track++;
		}
		//This is to skip letters we've already accounted for
		//If input is "aabb", you only want 1 loop for 'a' and 'b'
		i += track;
	}

	for (int i = 0; i < lengths.size(); i++)
		std::cout << lengths[i] << chars[i] << ' ';
}