leetcode--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].

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

The following code needs O(n) for the worst case.

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int[] indices = new int[]{-1,-1};
        int length = A.length;
        int index = Arrays.binarySearch(A, target);
        if(index >= 0){
        	int left = index, right = index;
        	while(left > -1 && A[left] == target){
                --left;
        	}
        	while(right < length && A[right] == target){
        	    ++right;
        	}
        	indices[0] = left + 1;
        	indices[1] = right - 1;
        }
        return indices;    
    }
}