One 、 subject

Two 、 Code

class Solution {
    
public:
    vector<int> searchRange(vector<int>& nums, int target) {
    
    
    int size= nums.size();
    
    //std::cout<<sz<<std::endl; 
    
    int left=0;
    int right=size-1;
    int middle=0;

    vector<int> return_vector;	   // The use of iterators , You have to use push_back() perhaps ={} initialization , Can only be 
    return_vector={
    -1,-1};
    
    int fist_dst=-1;   // The starting position 
    int last_dst=-1;   // Termination position 
    int middle_dst=-1;   // First found location  
    while(left<=right)
    {
    

        if(left<=right-2)      // Two points can be divided  
            {
            
                    middle=(left+right)/2;     // The questions are arranged in ascending order 

                    if(nums[middle]==target)    // Equal after two points   Return index value 
                    {
    
                    middle_dst=middle;
                    break;
                    } 
                    if(nums[middle]>target)  // The value after two points is larger than the target value   The middle value must not be   On the left 
                    {
    
                    right=middle-1;

                    }
                
                    if(nums[middle]<target)  // The value after two points is smaller than the target value   The middle value must not be   In zip code 
                    {
    
                        left=middle+1; 
                    
                    }
            }
        else    // Two cannot be separated   The two are equal or different 1
            {
    
                 if(nums[left]==target) 
                 {
    
                   middle_dst=left;
                   break;
                 }
                 if(nums[right]==target) 
                 {
    
                   middle_dst=right;
                   break;
                 }
                 if(nums[right]!=target&&nums[left]!=target)  break;              
            }
    }   
    // At this time, one coordinate has been found or none has been found  
    int i;

    std::cout<<"temp middle_dst" <<middle_dst<<std::endl;

    if(middle_dst!=-1)
    {
    
        
        for(i=middle_dst;i>=1;i--)   // Looking for the left boundary 
        {
    
           if(nums[i]==target&&nums[i-1]!=target)        
            {
    
             fist_dst=i;
             break;
            }
            if(nums[0]==target) 
             {
    
             fist_dst=0;
             break;
            }
        }

          for(i=middle_dst;i<=size-2;i++)   // Look for the right border 
        {
    
           if(nums[i]==target&&nums[i+1]!=target)        
            {
    
             last_dst=i;
             break;
            }
            if(nums[size-1]==target) 
             {
    
             last_dst=size-1;
             break;
            }
        }

        // As long as it can enter the cycle   We must be able to find the left and right boundaries   Next, consider the situation that you can't enter the left-right cycle 

       if(middle_dst==0) // Left boundary lookup   Leak filling 
       {
    
         if(nums[0]==target) fist_dst=0;

       }

       if(middle_dst>size-2)
       {
    
          if(nums[size-1]==target)  last_dst=size-1;

       }
    }

    return_vector[0]=fist_dst;
    return_vector[1]=last_dst;
    return return_vector;
    }
};

3、 ... and 、 Running results