lintcode61- Search for a Range- medium

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

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].

Challenge 

O(log n) time.

二分查找法找first target，然后向后检索。

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 int[] searchRange(int[] A, int target) {
        // find the first
        int[] result = new int[2];
        for (int i = 0; i < result.length; i++){
            result[i] = -1;
        }

        if (A == null || A.length == 0){
            return result;
        }

        int start = 0;
        int end = A.length - 1;
        while (start + 1 < end){
            int mid = start + (end - start) / 2;
            if (target <= A[mid]){
                end = mid;
            } else {
                start = mid;
            }
        }

        // search forward
        if (A[start] == target){
            result[0] = start;
            result[1] = start;
            while (result[1] + 1 < A.length && A[result[1] + 1] == target){
                result[1]++;
            }
        } else if (A[end] == target){
            result[0] = end;
            result[1] = end;
            while (result[1] + 1 < A.length && A[result[1] + 1] == target){
                result[1]++;
            }
        }
        return result;


    }
}