Adding/advancing is indeed easier. What about going backwards (insert before head or remove from tail), i.e when 0, then length. One way to do it is something like that, assume 2 compliment and there is negative shift left available. Not straightforward:

  int h = ( -(--head) >>> 31) ^ 1;//if head was equal to 1 (and only 1), h = 1, otherwise zero    
  head += h * len; //or shift left, if there is log2 len available; still, mul is a fast operation, unlike div

Of course, it can implemented via branching but that would be a major downside for 1-based idx.