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 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]

解題思路：

本題的解法是將vector中的每一個元素與target進行比對，當元素與target相等時將下標記錄下來。

C++程式碼：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) 
    {
        vector<int>res;
        vector<int>result;
        vector
 
<int>res_copy(2,-1);
        int len= nums.size();
        for(int i = 0;i<len;i++)
        {
            if(nums[i]==target)
            {
                res.push_back(i);
            }
        }
        if(res.size()==0)
        {
            return res_copy;
        }
        if(res.size()==1
 
)
        {
            res.push_back(res[0]);
            return res;
        }
        if(res.size()==2)
        {
            return res;
        }
        if(res.size()>2)
        {
            result.push_back(res[0]);
            result.push_back(res[res.size()-1]);
            return result;
        }
        return res;
    }
};

執行結果：