給定一個(gè)按照升序排列的整數(shù)數(shù)組 nums，和一個(gè)目標(biāo)值 target。找出給定目標(biāo)值在數(shù)組中的開始位置和結(jié)束位置。
你的算法時(shí)間復(fù)雜度必須是 O(log n) 級(jí)別。
如果數(shù)組中不存在目標(biāo)值，返回 [-1, -1]。

示例 1:
輸入: nums = [5,7,7,8,8,10], target = 8
輸出: [3,4]

示例 2:
輸入: nums = [5,7,7,8,8,10], target = 6
輸出: [-1,-1]

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] result = new int[2];
        int low = 0;
        int high = nums.length - 1;
        while(low <= high) {
            int mid = (low + high) / 2;
            if(nums[mid] == target) {
                int start = startPosition(nums, mid);
                int end = endPosition(nums, mid);
                result[0] = start;
                result[1] = end;
                return result;
            } else if(nums[mid] < target) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        result[0] = -1;
        result[1] = -1;
        return result;
    }
    
    private int startPosition(int[] nums, int curIndex) {
        int start = curIndex-1;
        while(start >= 0 && nums[start] == nums[curIndex]) {
            start--;
        }
        return start+1;
    }
    
    private int endPosition(int[] nums, int curIndex) {
        int end = curIndex+1;
        while(end < nums.length && nums[end] == nums[curIndex]) {
            end++;
        }
        return end-1;
    }
   
}