Build a simple regex function

Problem

Given a regular expression with characters a-z, ' * ', ' . '
the task was to find if that string could match another string with characters from: a-z
where ' * ' can delete the character before it, and ' . ' could match whatever character. ' * ' always appear after a a-z character.
Example:
isMatch("a*", "") = true;
isMatch(".", "") = false;
isMatch("ab*", "a") = true;
isMatch("a.", "ab") = true;
isMatch("a", "a") = true;

Solution

package myPackage;

public class Regex {

 boolean match(String r, String in)
 {
  int j= 0;
  for(int i= 0; i< r.length(); i++)
  {
   if(j >= in.length()) return false;
   if(r.charAt(i) != '.' && r.charAt(i) != '+' 
     && r.charAt(i) != '*' 
     && r.charAt(i) != '?')
   {
    int k = i+1;
    if((k< r.length() && r.charAt(k) != '.' 
      && r.charAt(k) != '+' 
      && r.charAt(k) != '*' 
      && r.charAt(k) != '?') 
      || k == r.length())
    {
     if(j< in.length() 
      && in.charAt(j)==r.charAt(i)) 
      j++;
     else return false;
     i--;
    }
    else if(r.charAt(k) == '*') 
     while(j< in.length() 
      && r.charAt(i) == in.charAt(j))
     {j++;}
    else if(r.charAt(k) == '?') 
     if(r.charAt(i) == in.charAt(j)) 
     {j++;}
    else if(r.charAt(k) == '+') 
     if(r.charAt(i) == in.charAt(j)) 
     {j++;} 
     else return false;
    else if(r.charAt(k) == '.') {
      if(r.charAt(i) == in.charAt(j)) 
      {j++;} 
      else return false;
      while(j< in.length() 
          && (r.charAt(i) == in.charAt(j)))
      {j++;}
     }
    i++;
   }
   else {
    int k = i;
    if(r.charAt(k) == '*') 
     while(j< in.length())
     {j++;}
    else if(r.charAt(k) == '?') 
      if(j< in.length()) 
       j++;
    else if(r.charAt(k) == '+') 
    {
     if(j< in.length()) j++; 
     else return false;
    }
    else if(r.charAt(k) == '.') 
    {
     if(j< in.length()) j++; 
     else return false;
     while(j< in.length()){j++;}
    }
   }
  }
  if(j< in.length()) return false;
  return true;
 }


 public static void main(String[] args) {

  Regex ob = new Regex();
  System.out.println(ob.match("aab*d*c*a*b*aa*", "aabbbcbbbaa"));
 }

}

Regex

Problem

How to find if string has alphanumeric characters?

Solution

Pattern p = Pattern.compile("[^0-9a-z]", Pattern.CASE_INSENSITIVE);

boolean b = p.matcher("My String").find();

Regex Lookup table

^	Matches the starting position within the string. In line-based tools, it matches the starting position of any line.
*	Matches the preceding element zero or more times. For example, ab*c matches "ac", "abc", "abbbc", etc. [xyz]* matches "", "x", "y", "z", "zx", "zyx", "xyzzy", and so on. (ab)* matches "", "ab", "abab", "ababab", and so on.
.	Matches any single character
?	Matches the preceding element zero or one time. For example, ab?c matches only "ac" or "abc".
+	Matches the preceding element one or more times. For example, ab+c matches "abc", "abbc", "abbbc", and so on, but not "ac".
|	The choice (also known as alternation or set union) operator matches either the expression before or the expression after the operator. For example,abc|def matches "abc" or "def".
{m,n}	Matches the preceding element at least m and not more than n times. For example, a{3,5} matches only "aaa", "aaaa", and "aaaaa".
[ ]	A bracket expression. Matches a single character that is contained within the brackets. For example, [abc] matches "a", "b", or "c".