一锭吨、問題描述
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:
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

二剃执、解決思路
思路一：題目要求時間復雜度為O(logn)，因此硅急，可以采用二分查找目標元素圣猎，查找到后采用分治分別求出目標元素的最小下標和最大下標逝钥，分治過程同樣是使用二分查找

三、算法實現
思路一

public int[] searchRange(int[] nums, int target) {
        int lens = nums.length;
        if(lens == 0) return new int[]{-1,-1};
        int left = 0;
        int right = lens - 1;
        int fidx = -1;
        int eidx = -1;
        // 先二分找出target下標
        int idx = binarySearch(nums, left, right, target);
        fidx = idx;
        eidx = idx;
        //System.out.println(fidx + ", " + eidx);
        if(fidx == -1) {
            return new int[]{-1,-1};
        }
        // 繼續(xù)二分查找最小下標
        if(fidx > 0){
            while(true){
                left = 0;
                right = fidx - 1;
                int min = binarySearch(nums, left, right, target);
                //System.out.println("min = " + min);
                if(min != -1){
                    fidx = min;
                } else {
                    break;
                }
            }
        }
        //System.out.println("============");
        // 二分查找最大下標
        if(eidx < lens - 1){
            while(true){
                left = eidx + 1;
                right = lens - 1;
                int max = binarySearch(nums, left, right, target);
                //System.out.println("max = " + max);
                if(max == -1){
                    break;
                } else {
                    eidx = max;
                }
            }
        }
        return new int[]{fidx,eidx};
    }

    public int binarySearch(int[] nums, int left, int right, int target){
        int idx = -1;
        while(left <= right){
            int mid = (left + right) / 2;
            if(nums[mid] == target) {
                idx = mid;
                break;
            } else if(nums[mid] > target){
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return idx;
    }