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CS111 -Decompress - Solved

Write your code in the file StringRec.java, your class must have this exact name with S and R capitalized.

Include a recursive method decompress with the following method header:

public static String decompress(String compressedText):
The following restrictions apply to method decompress:

YOUR CODE MUST BE RECURSIVE.
Do not use loops (while, do/while, or for).
Do not declare any variables outside of a method. You may declare local variables inside a method.
Method decompress will decompress text which has been compressed using the RLE algorithm defined in the previous assignment and described below.

Run-length encoding (RLE) is a simple "compression algorithm" (an algorithm which takes a block of data and reduces its size, producing a block 
that contains the same information in less space). It works by replacing repetitive sequences of identical data items with short "tokens" that 
represent entire sequences. Applying RLE to a string involves finding sequences in the string where the same character repeats. Each such sequence 
should be replaced by a "token" consisting of:

the number of characters in the sequence
the repeating character
If a character does not repeat, it appears as a single character in the compressed string with no number preceding it.

For example, consider the following string:

qwwwwwwwwweeeeerrtyyyyyqqqqwEErTTT
After applying the RLE algorithm, this string is converted into:

q9w5e2rt5y4qw2Er3T
In the compressed string, "9w" represents a sequence of 9 consecutive lowercase "w" characters. "5e" represents 5 consecutive lowercase "e" characters, etc.

For example, the following call to decompress

decompress("q9w5e2rt5y4qw2Er3T")
returns the string: "qwwwwwwwwweeeeerrtyyyyyqqqqwEErTTT"


For this problem, you may assume that the character counts will be single-digit numbers (a character will not repeat more than 9 times consecutively).

Hint #1: remember that characters are represented by numeric codes. You can decrement a character variable as follows:

 char c = '7';
 c--; // c will now hold the character '6'
 
Hint #2: You probably will not need to use this hint for this problem. However, a fast way to convert a digit character into the numeric value of the digit is to subtract the character code for the digit zero:

 char c = '7'; // this has the character code 55, not 7
 int x = c - '0'; // this produces the number 7
 

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