Find range of a target number

Given a sorted array with duplicates and a number, find the range in the 
form of (startIndex, endIndex) of that number. For example, 

find_range({0 2 3 3 3 10 10}, 3) should return (2,4). 
find_range({0 2 3 3 3 10 10}, 6) should return (-1,-1). 
The array and the number of duplicates can be large.

Click here for the original problem.
Now that I realize when dealing with searching problem, I should try to get an O(log(n)) solution. Yeah, binary search.

public class FindRange {
 public int[] findBound(int[] num, int target) {
  if (num == null)
   throw new NullPointerException("Null array");
  int[] bound = {-1, -1};
  if (num.length == 0)
   return bound;
  findRange(0, num.length - 1, bound, num, target);
  return bound;
  
 }
 private void findRange(int start, int end, int[] bound, int[] num, int target) {
  if (start > end)
   return;
  int mid = (start + end) / 2;
  if (num[mid] == target) {
   if (start == mid || num[mid - 1] != num[mid]) {
    if (bound[0] == -1 || mid < bound[0])
     bound[0] = mid;
   }
    
   else {
    findRange(start, mid - 1, bound, num, target);
   }
   if (end == mid || num[mid + 1] != num[mid]) {
    if (mid > bound[1])
     bound[1] = mid;
   }
   else {
    findRange(mid + 1, end, bound, num, target);
   }
  }
  else if (num[mid] < target)
   findRange(mid + 1, end, bound, num, target);
  else
   findRange(start, mid - 1, bound, num, target);
 }

}