It is necessary to repeat the character, as many times as the number behind it. 

They are positive integer numbers. 

```
case #1

input: "abc3leson11"

output: "abccclesonnnnnnnnnnn"

```

I already finish it in the following way:

```
    String a = "abbc2kd3ijkl40ggg2H5uu";
    String s = a + "*";
    String numS = "";
    int cnt = 0;
    for (int i = 0; i < s.length(); i++) {
        char ch = s.charAt(i);
        if (Character.isDigit(ch)) {
            numS = numS + ch;
            cnt++;
        } else {
            cnt++;
            try {
                for (int j = 0; j < Integer.parseInt(numS); j++) {
                    System.out.print(s.charAt(i - cnt));
                }
                if (i != s.length() - 1 && !Character.isDigit(s.charAt(i + 1))) {
                    System.out.print(s.charAt(i)); 
                }
            } catch (Exception e) {
                if (i != s.length() - 1 && !Character.isDigit(s.charAt(i + 1))) {
                    System.out.print(s.charAt(i)); 
                }
            }
            cnt = 0;
            numS = "";
        }

    }

```

But I wonder is there some better solution with less and cleaner code?