力扣算法题—149Max Points on a line

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

Example 1:

Input: [[1,1],[2,2],[3,3]]
Output: 3
Explanation:
^
|
|        o
|     o
|  o  
+------------->
0  1  2  3  4

Example 2:

Input: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
Output: 4
Explanation:
^
|
|  o
|     o        o
|        o
|  o        o
+------------------->
0  1  2  3  4  5  6

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

Solution：

　　直接求每个点与其他点的斜率，然后计算相同斜率的个数就行

　　注意两点：

　　一个是会有相同的点出现，另一个是斜率计算得到无穷和double比较相等的情况

 class Solution {
 public:
     int maxPoints(vector<vector<int>>& points) {
         if (points.size() < )return points.size();
         int res = ;
         for (int i = ; i < points.size(); ++i)
         {
             int same = ;
             for (int j = i + ; j < points.size(); ++j)
             {
                 int cnt = ;
                 long long int x1 = points[i][], x2 = points[j][];
                 long long int y1 = points[i][], y2 = points[j][];
                 if (x1 == x2 && y1 == y2) { ++same; continue; }
                 for (int k = ; k < points.size(); ++k)
                 {
                     long long int x3 = points[k][], y3 = points[k][];
                     if ((x1*y2 + x2 * y3 + x3 * y1 - x3 * y2 - x2 * y1 - x1 * y3) == )
                         ++cnt;
                 }
                 res = max(res, cnt);
             }
             res = max(res, same);
         }
         return res;
     }
 };