basic java program not working

Question

For the code i need to write a method the decompresses a string. For example if the user entered "2d5t" the method would return "ddttttt". My code now will work for that input but if the input uses a character without a number before it the program wont work when it should. For example if the input was just "d" the program wouldnt work instead of just returning "d". The code also has to be recursive. Here is what my code is now please help.

public static String decompress(String compressedText) {
    if (compressedText.equals(""))
        return "";
    return decompress(compressedText, charInt(compressedText, 0), 0);
}

public static String decompress(String text, int count, int pos) {
    if (pos == text.length() || (pos == text.length()-2 && count == 0))
        return "";
    else if (count == 0)
        return decompress(text, charInt(text, pos+2), pos+2);
    return text.charAt(pos+1) + decompress(text, count-1, pos);
}

public static int charInt(String str, int idex) {
    return str.charAt(idex) - '0';
}

Do you use an escape character? Somebody might want to compress "1112223333344". — Eric Jablow
– Eric Jablow, Commented Nov 28, 2013 at 2:37

Nathan Merrill · Accepted Answer · 2013-11-28 03:03:42Z

1

Here's some pseudocode:

function createString(int times, char character){
    if times is 0, do nothing
    otherwise return character + createString(times-1, character);
}

function createString(string full){
    split string by number/character pairs
    for each pair, call createString(times, character), and append them
}

I don't believe in handing out real code, sorry. It's much better in the long run.

answered Nov 28, 2013 at 3:03

Nathan Merrill

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Comments

Muhammad Maqsoodur Rehman · Accepted Answer · 2013-11-28 02:37:34Z

0

You need to validate your user input. Decide first, that what string values are acceptable to your method and then write a validate method. Then invoke that method inside your decompress method. Look into string manipulation functions and regular expressions in Java. And then try rewriting your code.

answered Nov 28, 2013 at 2:37

Muhammad Maqsoodur Rehman

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cosjav · Accepted Answer · 2013-11-28 02:55:32Z

0

As mentioned by others, this can be solved with regular expressions. An example solution is:

  public static String decompress(String compressed) {
    Matcher matcher = Pattern.compile("(\\d+)([^\\d])").matcher(compressed);
    StringBuffer decompressed = new StringBuffer();
    while (matcher.find()) {
      Integer charNum = Integer.parseInt(matcher.group(1));
      StringBuilder decompressedChars = new StringBuilder();
      for (int i = 1; i <= charNum; i++) {
        decompressedChars.append(matcher.group(2));
      }
      matcher.appendReplacement(decompressed, decompressedChars.toString());
    }
    matcher.appendTail(decompressed);
    return decompressed.toString();
  }

This code won't support numbers larger than Integer.MAX_VALUE and you might want to put some error handling and validation in there also.

answered Nov 28, 2013 at 2:55

cosjav

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4 Comments

cosjav Over a year ago

Is this for an assignment or something? Because thats a strange requirement otherwise...

user3044112 Over a year ago

yea this is an assignment for my intro to compsci class

user3044112 Over a year ago

Do you not see the code i already did? im just stuck with an error i obviously already tried

cosjav Over a year ago

I provided this answer before you edited your question to include "The code also has to be recursive". Now that you mentioned this is for an assignment, I agree with MrTi.

Jules Gagnon-Marchand · Accepted Answer · 2013-11-28 04:39:35Z

**Edited to be recursive as per the OP's request

Tested left first one char lookahead parser using regex

import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class Parser{

private static String parse(StringBuilder output, String input, Integer offset){
    if(offset<input.length()){
        java.util.regex.Pattern p0 = 
            java.util.regex.Pattern.compile("\\d(?=[a-z])");

        java.util.regex.Pattern p1 = 
            java.util.regex.Pattern.compile("[a-z]");

        java.util.regex.Matcher m0 = p0.matcher(input);
        java.util.regex.Matcher m1 = p1.matcher(input);
        if (m0.find(offset) && m0.start() == offset) 
        {
            for(Integer i = 0;
                i < Integer.parseInt(String.valueOf(input.charAt(offset))); 
                ++i) {
               output.append(input.charAt(offset+1));
            }
            offset+=2;
        }  
        else if (m1.find(offset) && m1.start() == offset) {
            output.append(input.charAt(offset));
            ++offset;
        }
        else {
            ++offset;
        }
        return parse(output, input, offset);
    }
    else return output.toString();
}

public static void main(String[] args) 
{
    Integer offset = 0;
    StringBuilder output = new StringBuilder();
    parse(output, args[0], offset);
    System.out.println(output.toString());    
}
}

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