Recursion coding bat Given a non-negative int n, compute recursively (no loops) the count of the occurrences of 8 as a digit, except that an 8 with another 8 immediately to its left counts double

Given a non-negative int n, compute recursively (no loops) the count of the occurrences of 8 as a digit, except that an 8 with another 8 immediately to its left counts double, so 8818 yields 4. Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).

count8(8) → 1
count8(818) → 2
count8(8818) → 4

public int count8(int n) {
  
    if(n==0)
    {
      return 0;
    }
    else if(n%10==8)
    {
      int temp = n/10;
      if(temp%10==8)
      {
        return (2+ count8(n/10));
      }
    
      return (1+ count8(n/10));
      
    }
    
    else 
    {
      return count8(n/10);
    }

}