Simple Regex Parsing

gistfile1.cs

/*Implement a function which answers whether a given string matches a provided pattern. The pattern is written in a small regex-like language, consisting of:

- Lowercase latin characters (a to z)
- . (dot), which matches any latin character.
- *, indicates that the previous character is repeated 0 or more times.
- +, indicates that the previous character is repeated 1 or more times.

You can assume that the input pattern is well formed and the string to match consists of lowercase latin characters only.

Examples:

match('abba', 'abba') == true
match('cda', 'adb) == false
match('.*abc', 'abc') == true
match('.*abc', 'ababc') == true
match('.b*ac*aa+', 'caaa') == true
match('bb+a*', 'baa') == false

match('.*abc.*a*b*', 'ababc') == true

'.*abc' -> ['.*', 'a', 'b', 'c']
*/

public bool match(string regex, string input) {
    string newRegex = "";
    for (int i = 0; i < regex.length; i++) {
        if (regex[i] == "+") {
            newRegex+= regex[i-1]+"*";
        }
        else
        {
            newRegex+=regex[i];
        }
    }
    
    return rHelper(newRegex, input);
}

public bool rHelper(string regex, string input) {
    if (input == "") {
        for (int i = regex.length -1; i >= 0; i--) {
            if (regex[i] == "*") {
                i--;
            }
            else
            {
                return false;
            }
        }
        
        return true;
    }
    
    if (regex[0] == input[0] || regex[0] == ".") {
       if (regex.length > 1 && regex[1] == "*") {  // possible bug (in case where regex.length == 1)
            bool r1 = rHelper(regex, input.substring(1));
            bool r2 = rHelper(regex.substring(2), input);
            
            return r1 || r2;
       }
       else
       {
           return rHelper(regex.substring(1), input.substring(1));
       }  
    }
    
    return false;
}