LeetCode：149_Max Points on a line | 寻找一条直线上最多点的数量 | Hard

题目：Max Points on a line

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

这道题需要稍微转变一下思路，用斜率来实现，试想找在同一条直线上的点，怎么判断在一条直线上，唯一的方式也只有斜率可以完成，我开始没想到，后来看网友的思路才想到的，下面是简单的实现：
其中有一点小技巧，利用map<double, int>来存储具有相同斜率值的的点的数量，简洁高效。

 /*
     Definition for a point
 */
 // struct Point {
 //     int x;
 //     int y;
 //     Point():x(0),y(0) {}
 //     Point (int a, int b):x(0), y(0) {}
 // };

 int maxPoints(vector<Point> &points) {
     if (points.empty())
         return ;
     if (points.size() <= )
         return points.size();

     int numPoints = points.size();
     map<double, int> pmap;    //存储斜率-点数对应值
     int numMaxPoints = ;

     for (int i = ; i != numPoints - ; ++i) {
         int numSamePoints = , numVerPoints = ;  //针对每个点分别做处理

         pmap.clear();

         for (int j = i + ; j != numPoints; ++j) {
             if(points[i].x != points[j].x) {
                 double slope = (double)(points[j].y - points[i].y) / (points[j].x - points[i].x);
                 if (pmap.find(slope) != pmap.end())
                     ++ pmap[slope];   //具有相同斜率值的点数累加
                 else
                     pmap[slope] = ;
             }
             else if (points[i].y == points[j].y)
                 numSamePoints ++;   //重合的点
             else
                 numVerPoints ++;    //垂直的点

         }
         map<double, int>::iterator it = pmap.begin();
         for (; it != pmap.end(); ++ it) {
             if (it->second > numVerPoints)
                 numVerPoints = it->second;
         }
         if (numVerPoints + numSamePoints > numMaxPoints)
             numMaxPoints = numVerPoints + numSamePoints;
     }
     return numMaxPoints + ;
 }