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;
    }
}