All geek questions in one place

Recursive decompression of strings

I am trying to unzip strings that look like this:

Entrance: 4 (ab)

Output: abababab

Entrance: 11ab

Output: aaaaaaaaaaab

Entrance: 2 (3b3 (ab))

Output: bbbabababbbbabababab

The above examples are output correctly using the recursive method below, but problems arise when I enter something like:

Entrance: 4 (ab) a

Expected Result: ababababa

Entrance: 2 (3b3 (ab)) a

Expected Result: bbbabababbbbabababa

, , "return repeat". , . , , , , - . 2 (3b3 (ab)) a 2 * (3b3 (ab)) + a, 2 * (3b3 (ab)) a. , .

public static String decompress(String compressedText) throws Exception
{
   //BASE CASE 
    if(compressedText.length() == 1)
    {
        if(compressedText.charAt(0) == ')')
        {
            System.out.println("1: " + compressedText);
            return "";
        }
        else
        {
            System.out.println("2: " + compressedText);
            return compressedText;
        }

    }
    //END BASECASE


    if(compressedText.charAt(0) == '(')
    {
        System.out.println("3: " + compressedText);
        return decompress(compressedText.substring(1));        
    }


    //IF DOUBLE DIGIT
    if(Character.isDigit(compressedText.charAt(0)) == true && Character.isDigit(compressedText.charAt(1)) == true)
    {
        if(compressedText.charAt(3) != '(')
        {
            System.out.println("4: " + compressedText);
            int i = Integer.parseInt(compressedText.substring(0,2));
            String repeated = new String(new char[i]).replace("\0", compressedText.substring(2,3));  
            return repeated + decompress(compressedText.substring(3));
        }
        else
        {
            System.out.println("5: " + compressedText);
            int i = Integer.parseInt(compressedText.substring(0,2));
            String repeated = new String(new char[i]).replace("\0", decompress(compressedText.substring(2)));
            return repeated;
        }

    }
    //END DOUBLE DIGIT



    //IF SINGLE DIGIT
    if (Character.isDigit(compressedText.charAt(0)) == true)
    {
        if(compressedText.charAt(1) !='(')
        {
            System.out.println("6: " + compressedText);
            int i = Integer.parseInt(compressedText.substring(0,1));
            String repeated = new String(new char[i]).replace("\0", compressedText.substring(1,2));  
            return repeated + decompress(compressedText.substring(2)); 
        }
        else
        {
            System.out.println("7: " + compressedText);
            int i = Integer.parseInt(compressedText.substring(0,1));
            String repeated = new String(new char[i]).replace("\0", decompress(compressedText.substring(1)));
            return repeated;
        }

    }
    //END SINGLE DIGIT

    //IF RIGHT PARENTHESIS
    if (compressedText.charAt(0) == ')')
    {
        if (compressedText.charAt(1) != ')')
        {
            System.out.println("8: " + compressedText);
            return "";
        }
        else
        {
            System.out.println("9: " + compressedText);
            return  decompress(compressedText.substring(1));

        }

    }
    //END 

        System.out.println("10: " + compressedText);
        return compressedText.charAt(0)+decompress(compressedText.substring(1));

}
+4
3

, , , "" "a", "" "8:". , - , - bbbabaabaababbbabaabaaba.

, , ( , , , ...).

, , . , . : , , .. ( , .)

, ( char, , ). , char .

public static String decompressWithRegex(String s) {
    if ((s == null) || (s.length() == 0)) {
        return s;
    }
    // pattern for finding number with either bracket-enclosed, char-only part or single char
    Pattern p = Pattern.compile("(\\d+)((?:[^\\d\\(\\)]{1})|(?:\\([^\\d\\(\\)]+\\)))");
    String tmp = s;
    Matcher m = p.matcher(tmp);
    // start searching
    while (m.find(0)) {
        // first capture group returns count
        int count = Integer.parseInt(m.group(1));
        // second group is string to repeat (if it bracket-enclosed, then remove brackets)
        String what = m.group(2).replace("(", "").replace(")", "");
        // build replacement part
        String replacePart = repeat(what, count);
        // replace it
        tmp = m.replaceFirst(replacePart);
        // reset matcher (source of matcher is now the new string)
        m.reset(tmp);
    }
    return tmp;
}

. , :

  • , , ,
  • , ,
  • , ,
  • , .
  • , , .
  • , , , .
  • ,

:

public static String decompressWithSearching(String s) {
    if ((s == null) || (s.length() == 0)) {
        return s;
    }
    // replace non-groups first
    for (int i = s.length() - 1; i >= 0; i--) {
        // find digit that is not followed by bracket
        if (Character.isDigit(s.charAt(i)) && s.charAt(i + 1) != '(') {
            // string to repeat is right behind the digit
            String part = s.substring(i + 1, i + 2);
            // find complete digit
            String countStr = "";
            int j = i;
            for ( ; j >= 0 && Character.isDigit(s.charAt(j)); j--) {
                countStr = s.charAt(j) + countStr;
            }
            int count = Integer.parseInt(countStr);
            // build replacement part
            String replacePart = repeat(part, count);
            // replace part
            s = s.substring(0, j + 1) + replacePart + s.substring(i + 2);
        }
    }

    // replace nested parts
    int closing;
    while ((closing = s.indexOf(')')) > -1) {
        // find matching opening bracket
        int opening = s.lastIndexOf('(', closing);
        // text between is to be repeated
        String what = s.substring(opening + 1,closing);
        // find complete digit
        String countStr = "";
        int numPartIndex = opening - 1;
        while (numPartIndex >= 0 && Character.isDigit(s.charAt(numPartIndex))) {
            countStr = s.charAt(numPartIndex) + countStr;
            numPartIndex--;
        }
        int count = Integer.parseInt(countStr);
        // build replacement part
        String replacePart = repeat(what, count);
        // replace part
        s = s.substring(0, numPartIndex + 1) + replacePart + s.substring(closing + 1);
    }

    return s;
}

:

public static String repeat(String what, int times) {
    if ((times <= 0) || (what == null) || (what.length() == 0)) {
        return "";
    }
    StringBuilder buffer = new StringBuilder(times * what.length());
    for (int i = 0; i < times; i++) {
        buffer.append(what);
    }
    return buffer.toString();
}
+1

, :

index 0 1 2 3 4 5 6 7 8 9 10
str   2 ( 3 b 3 ( a b ) ) a

  f(0)

  => 2 * f(1)[0] add f(f(1)[1] + 1)  // f(1)[1] is the closing index 

    f(1) => 3 * b + 3 * f(5)[0] add f(f(5)[1] + 1)

    => f(5) returns (ab,8)

    f(1) => bbb + ababab add f(9) // str[9] is closing parenthesis

    => f(1) returns (bbbababab,9)

  => 2 * bbbababab add f(10)

  => bbbabababbbbabababa

JavaScript:

var example = '2(3b3(ab)2(cd3(fg)))ab2(gh2(xz))';

console.log(example);
console.log(decompress(example));

function decompress(s){

  // returns tuple [accumulator, index of closing parenthesis]
  function f(i){
  
    var accum = '',
        mult = '',
        curr = '';
      
    // accumulate all parenthetical groups in this level  
    while (i !== s.length){

      // closing parenthesis
      if (s[i] === ')'){
      
        // add the last decompression
        if (curr !== ''){
          accum += customReplicate(curr,mult);
        }
        
        // exit this call
        return [accum,i];
      }
          
      // character is a digit
      if (!isNaN(parseInt(s[i]))){
      
        // add previous decompression
        if (curr !== ''){
          accum += customReplicate(curr,mult);
          
          curr = '';
          mult = s[i];
          
        } else {
          mult += s[i];
        }
        
        i++;
        
      // character is a character
      } else if (s[i] !== '('){
      
        curr += s[i];
        i++;
        
      // parenthetical group 
      } else if (s[i] === '('){
      
        // recursive call
        [tempAccum,index] = f(i + 1);

        accum += customReplicate(tempAccum,mult);
        mult = '';
        i = index + 1;
      }
    }
    
    return accum + customReplicate(curr,mult);
  }
  
  // initialize the recursion
  return f(0);
}

function customReplicate(str,times){
  return new Array(times === '' ? 1 : parseInt(times))
                 .fill(str).join('');
}
Hide result
+1

I understand that this is a Java issue, but I usually write a little Ruby code to test the idea before implementing it in Java. If he is interested in someone, here is my code:

def decompress(str)
  str.gsub!(/(\d+)([a-z])/i){$2*$1.to_i}       # Replace every subtring like "3b" and "11a".
  while str.include?('(') do
    str.sub!(/(\d+)\(([a-z]+)\)/){$2*$1.to_i}  # Replace the first inner group found
  end
  str
end

puts decompress("4(ab)")       == "abababab"
puts decompress("11ab")        == "aaaaaaaaaaab"
puts decompress("2(3b3(ab))")  == "bbbabababbbbababab"
puts decompress("4(ab)a")      == "ababababa"
puts decompress("2(3b3(ab))a") == "bbbabababbbbabababa"
#=> true, true, true, true, true

@jCoder wrote almost the same thing in his first example, so no need to reinvent the wheel!

0
source

Source: https://habr.com/ru/post/1662735/

More articles:


All Articles