class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int left=0,right=nums.size()-1,mid;
        while(left<=right){
            mid=(left+right)/2;
            
            if(target==nums[mid]){
                int i=mid,j=mid;
                while
 
(target==nums[i-1]&&i>0) i--;
                while(target==nums[j+1]&&j<nums.size()-1) j++;
                return {i,j};
            }
            else if(target>nums[mid]) left=mid+1;
            else right=mid-1;
        }
        return {-1,-1};
    }
};

上面的演算法不是嚴格意義上的O(logn)的演算法，因為在最壞的情況下會變成O(n)，比如當數組裡的數全是目標值的話，從中間向兩邊找邊界就會一直遍歷完整個陣列，那麼我們下面來看一種真正意義上的O(logn)的演算法，使用兩次二分查詢法，第一次找到左邊界，第二次呼叫找到右邊界即可，具體程式碼如下：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        if(nums.empty()) return {-1,-1};
        vector<int> res(2, -1);
        int left = 0, right = nums.size() - 1;
        while (left < right) {
            int mid = left + (right -
 
 left) / 2;
            if (nums[mid] < target) left = mid + 1;
            else right = mid;
        }
        if (nums[right] != target) return res;
        res[0] = right;
        right = nums.size()-1;
        while (left < right) {
            int mid =(right + left) / 2+1;//整型除法只會向下取整，現改為向上取整，保證中間值一直>left，可以一直向右檢查
            if (nums[mid] == target) left = mid;//只存在>或==的情況
            else right= mid-1;
        }
        res[1] = left;
        return res;
    }
};