LintCode-Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

Example

Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Solution:

 public class Solution {
     /**
      *@param A : an integer sorted array
      *@param target :  an integer to be inserted
      *return : a list of length 2, [index1, index2]
      */
     public ArrayList<Integer> searchRange(ArrayList<Integer> A, int target) {
         ArrayList<Integer> res = new ArrayList<Integer>();
         int start = -1, end = -1;
         int p1 = 0, p2 = A.size()-1;
         //find start point.
         while (p1<=p2){
             int mid = (p1+p2)/2;

             if (A.get(mid)==target){
                 if (mid==0 || A.get(mid-1)!=target){
                     start = mid;
                     break;
                 } else {
                     p2 = mid-1;
                 }
             } else if (A.get(mid)>target)
                 p2 = mid-1;
             else p1 = mid+1;
         }

         //find end point.
         p1 = 0;
         p2 = A.size()-1;
         while (p1<=p2){
             int mid = (p1+p2)/2;

             if (A.get(mid)==target){
                 if (mid==A.size()-1 || A.get(mid+1)!=target){
                     end = mid;
                     break;
                 } else p1 = mid+1;
             } else if (A.get(mid)>target)
                 p2 = mid-1;
             else p1 = mid+1;
         }

         res.add(start);
         res.add(end);

         return res;
     }
 }