Given an array of integers nums 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].

Example 1:
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

难度：medium

题目：
给定一升序整数数组，找出给定整数的起止边界。算法时间复杂度要为O(log n)

思路：二叉搜索

Runtime: 4 ms, faster than 65.40% of Java online submissions for Find First and Last Position of Element in Sorted Array.
Memory Usage: 30.5 MB, less than 28.29% of Java online submissions for Find First and Last Position of Element in Sorted Array.

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] result = new int[2];
        result[0] = binarySearch(nums, false, target);
        result[1] = binarySearch(nums, true, target);
        
        return result; 
    }
    // false -> left, true -> right
    private int binarySearch(int[] nums, boolean direction, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                if (direction && mid < nums.length - 1 && nums[mid + 1] == target) {
                    left = mid + 1;
                } else if (!direction && mid > 0 && nums[mid - 1] == target) {
                    right = mid - 1;
                } else {
                    return mid;
                }
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return -1;
    }
}