問題描述：

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(logn).
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].

問題來自此處

解答思路：
一看到題目要求時間複雜度是O(logn)，又是跟查詢相關的，還能想到什麼辦法呢~當然是二分法啦！
我的思路是先找到目標數，然後往目標數的左右分別延伸查詢，定下相同數所在的上下界。

vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> result(2,-1); //初始化為[-1,-1]
        int size = nums.size();
        if
 
(size == 0){
            return result;
        }
        int low = 0, high = size - 1, med;
        while(low <= high){
            med = (low + high) / 2;
            if(nums[med] == target){
                int left = find_bound(nums,0,med,target);
                int right = find_bound(nums,size,med,target);
                result.clear();
                result.push_back(left);
                result.push_back(right);
                return
 
 result;
            }
            else if(nums[med] < target){
                low = med + 1;
            }
            else{
                high = med - 1;
            }
        }
        return result;
    }
    int find_bound(vector<int> nums,int numb,int med,int target){
        int res = med;
        if(numb<=med){
            for(int i = med;i >= numb;i--){
                if(nums[i] == target)
                    res = i;
                if(nums[i] < target)
                    break;
            }
        }
        else{
            for(int i = med;i < numb;i++){
                if(nums[i] == target)
                    res = i;
                if(nums[i] > target)
                    break;
            }
        }
        return res;
    }

寫得很煞筆。。我以為執行時間的排名鐵定會被排在最後那批。。結果居然在中間……有毒有毒~