149. 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.

要求是输入n个坐标点，输出最多有多少个点在同一条直线上。

代码提交如下：

 1 int Solution::maxPoints(vector<Point>& points)
 2 {
 3     if (points.size()<=2)
 4     {
 5         return points.size();
 6     }
 7
 8     unsigned int max_line_nums = 0;
 9
10     for (unsigned int i = 0;i < points.size(); i++)
11     {
12         slope_map same_line;
13         unsigned int same_line_nums = 0,
14                      same_x_nums = 0,
15                      same_y_nums = 0,
16                      same_points_nums = 0;
17         for (unsigned int  j = i + 1;  j < points.size(); j++)
18         {
19             if (points[i].x == points[j].x && points[i].y == points[j].y)
20             {
21                 ++same_points_nums;
22             }
23             else if (points[i].x == points[j].x)
24             {
25                 ++same_x_nums;
26             }
27             else if (points[i].y == points[j].y)
28             {
29                 ++same_y_nums;
30             }
31             else
32             {
33                 double slope = ((double)points[i].y - (double)points[j].y)/((double)points[i].x - (double)points[j].x);
34                 double b = points[i].y - slope * points[i].x;
35
36                 same_line[slope][b]++;
37                 same_line_nums = (same_line_nums < same_line[slope][b])?same_line[slope][b]:same_line_nums;
38             }
39         }
40         unsigned int tmp_max_line_nums = same_points_nums + max(same_line_nums, (same_x_nums < same_y_nums)?same_y_nums:same_x_nums);
41         max_line_nums =  tmp_max_line_nums < max_line_nums ? max_line_nums:tmp_max_line_nums;
42     }
43
44     return max_line_nums + 1;
45 }

不足之处：

利用y = kx + b 可以将k和b存在一个map中，但是对于很大的数来说，精度会引起误差，使得两个不同的点被认为是同一个点。