Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

【thought】Use two layers of loop to solve it.Do not start the second loop from points.begin() but item1+1 (item1 is the current item of the first loop).We can avoid accessing any 2 duplicated points which could compose one straight line with this method.Any better solution?If any there,pls tell me.Thanks.

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6  7 using namespace std; 8  9 /**10  * Definition for a point.11  * struct Point {12  *     int x;13  *     int y;14  *     Point() : x(0), y(0) {}15  *     Point(int a, int b) : x(a), y(b) {}16  * };17  */18  19 bool typeisless(const Point &data1,const Point &data2){20     return data1.x < data2.x;21 }22  23 class Solution {24 public:25 26     int maxPoints(vector
          
            &points) {27 if(points.size()==0) return 0;28 29 int last_x = (points.begin()->x) - 1;30 int maxnum = INT_MIN,curnum = 0;31 std::sort(points.begin(),points.end(),typeisless);32 for(vector
           
            ::iterator iter = points.begin();33 iter != points.end();++iter){34 if(iter->x == last_x){35 ++curnum;36 }else{37 last_x = iter->x;38 curnum = 1;39 }40 41 maxnum = maxnum
            
              count_map;45 count_map.insert(pair
             
              (0.0,0));46 47 for(vector
              
               ::iterator iter1 = points.begin();48 iter1 != points.end();++iter1){49 count_map.clear();50 int samepointnum = 0;51 52 for(vector
               
                ::iterator iter2 = iter1 + 1;53 iter2 != points.end();++iter2){54 55 if(iter2->x == iter1->x && iter2->y == iter1->y){56 ++samepointnum;continue;57 }else if(iter2->x == iter1->x)58 continue;59 else{60 double dslope = (static_cast
                
                 (iter2->y - iter1->y))/61 (static_cast
                 
                  (iter2->x - iter1->x));62 if(count_map.find(dslope) != count_map.end()) ++count_map[dslope];63 else{64 count_map[dslope] = 1;65 ++count_map[dslope];66 }67 }68 }69 70 for(unordered_map
                  
                   ::iterator iter=count_map.begin();71 iter!=count_map.end();++iter){72 if(iter->second + samepointnum > maxnum){73 maxnum = iter->second + samepointnum;74 }75 }76 }77 78 return maxnum;79 }80 };