34. Search for a Range

Given an array of integers sorted in ascending order, 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].

给定一个整型升序数组，找到一个给定值在数组中的位置区间。 你的算法复杂度必须低于O(logn)。 如果目标在数组中不会被发现，返回[-1, -1]。 例如，给定[5, 7, 7, 8, 8, 10]，目标值为8，返回[3, 4]。

    public int[] searchRange(int[] nums, int target) {
        int[] result = {-1,-1};
        if (nums.length == 0) return result;
        int low = 0, high = nums.length - 1;
        while (low < high) {
            int mid = (low + high) / 2;
            if (nums[mid] < target) low = mid + 1;
            else  high = mid;
        }
        if (nums[low]!=target) return result;
        else result[0]=low;

        high = nums.length-1;  // We don't have to set i to 0 the second time.
        while (low < high)
        {
            int mid = (low + high) /2 + 1;    // Make mid biased to the right
            if (nums[mid] > target) high = mid - 1;
            else low = mid;                // So that this won't make the search range stuck.
        }
        result[1] = high;
        return result;       
    }