LeetCode：Max Points on a line

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。

 1 /* 

 2     Definition for a point

 3 */

 4 // struct Point {

 5 //     int x;

 6 //     int y;

 7 //     Point():x(0),y(0) {}

 8 //     Point (int a, int b):x(0), y(0) {}

 9 // };

10 

11 int maxPoints(vector<Point> &points) {

12     if (points.empty())

13         return 0;

14     if (points.size() <= 2) 

15         return points.size();

16 

17     int numPoints = points.size();

18     map<double, int> pmap;    //存储斜率-点数对应值

19     int numMaxPoints = 0;

20 

21     for (int i = 0; i != numPoints - 1; ++i) {

22         int numSamePoints = 0, numVerPoints = 0;  //针对每个点分别做处理

23         

24         pmap.clear();

25         

26         for (int j = i + 1; j != numPoints; ++j) {

27             if(points[i].x != points[j].x) {

28                 double slope = (double)(points[j].y - points[i].y) / (points[j].x - points[i].x);

29                 if (pmap.find(slope) != pmap.end())

30                     ++ pmap[slope];   //具有相同斜率值的点数累加

31                 else 

32                     pmap[slope] = 1;

33             }

34             else if (points[i].y == points[j].y) 

35                 numSamePoints ++;   //重合的点

36             else

37                 numVerPoints ++;    //垂直的点

38 

39         }

40         map<double, int>::iterator it = pmap.begin();

41         for (; it != pmap.end(); ++ it) {

42             if (it->second > numVerPoints)

43                 numVerPoints = it->second;

44         }

45         if (numVerPoints + numSamePoints > numMaxPoints) 

46             numMaxPoints = numVerPoints + numSamePoints;

47     }

48     return numMaxPoints + 1;

49 }