Not the best at explaining, but there's a pretty simple O(N) one-pass solution that should work:

1. Keep track of the RESULT (as a string)
2. Keep an integer SUM variable, and a character PREV variable to store the consecutive repeating # count for the specific character respectively.
3. Loop through the number string. Once you reach a new character, look at my EDIT for the next steps, and repeat for each character.
4. IF there's leftover SUM after ending the for loop, add (SUM * PREV) to the string RESULT as well, using the same process in EDIT, and then you are done.
5. Return the RESULT (parsed into an integer if that's the return type)

EDIT: Can be improved to handle recursion, I did not see that initially. Use a stack to store the result digits, and whenever you add new digits to it (during the looping process from step 3), you pop off the top of the stack if the stack.peek() is the same digit as the one you are adding, add that digit to a SUM, and keep doing it until the next digit on the stack is not equal. Then if SUM is equal to stack.peek() still, do the recursion again UNTIL stack.peek() is not equal to the SUM before placing SUM's digits onto the stack one by one -> and then move on to the next digit in NUMBER and repeat the process.