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].

使用二分查找，找到之后左右移动，获取边界。

    public int[] searchRange(int[] A, int target) {
        int len = A.length;
        int left = 0;
        int right = len - 1;
        

        int flag = -1;
        while (left <= right) {
        	
            int median = left + (right - left) / 2;
            if (A[median] == target) {
            	flag = median;
            	break;
            } else if (A[median] > target) {
                right = median - 1;
            } else {
                left = median + 1;
            }
        }
        int[] result;
        if (flag != -1) {
            int leftIndex = flag;
            int rightIndex = flag;
            while (rightIndex < len) {
            	if (A[rightIndex] == target) {
            		rightIndex++;
            	} else {
            		break;
            	}            	
            }
            while (leftIndex >= 0) {
            	if (A[leftIndex] == target) {
            		leftIndex--;
            	} else {
            		break;
            	}
            	
            }
            result = new int[]{leftIndex + 1, rightIndex - 1};
        } else {
            result = new int[]{-1, -1};
        }
        return result;
    }