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].

题目：给定一个排好序的整数数组，找到给定目标值的出现的首尾位置。

思路：二分查找。由于是有序数组，所以相同值的数是连续的，即只要找到其中一个，再向左右找到边界值就可以了，这三步均采用二分查找。

	public int[] searchRange(int[] A, int target) {
		int result[] = new int[] { -1, -1 };
		int len = A.length;
		if (len <= 0)
			return result;
		int separator = -1;
		int left = 0, right = len - 1;
		while (left <= right) {
			int middle = (left + right) / 2;
			if (A[middle] > target)
				right = middle - 1;
			else if (A[middle] < target)
				left = middle + 1;
			else {
				separator = middle;
				break;
			}
		}
		if (separator == -1)
			return result;
//找到左边界
		left = 0;
		right = separator;
		while (left <= right) {
			int middle = (left + right) / 2;
			if (A[middle] == target)
				right = middle - 1;
			else
				left = middle + 1;
		}
		result[0] = left;
//找到右边界
		left = separator;
		right = len - 1;
		while (left <= right) {
			int middle = (left + right) / 2;
			if (A[middle] == target)
				left = middle + 1;
			else
				right = middle - 1;
		}
		result[1] = right;
		return result;
	}

reference : http://www.darrensunny.me/leetcode-search-for-a-range/