[LeetCode] Max Points on a Line, Solution

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
[Thoughts]
任意一条直线都可以表述为
y = ax + b
假设，有两个点(x1,y1), (x2,y2)，如果他们都在这条直线上则有
y1 = kx1 +b
y2 = kx2 +b
由此可以得到关系，k = (y2-y1)/(x2-x1)。即如果点c和点a的斜率为k, 而点b和点a的斜率也为k，那么由传递性，可以知道点c和点b也在一条线上。解法就从这里来
取定一个点(x_k,y_k), 遍历所有节点(x_i, y_i), 然后统计斜率相同的点数，并求取最大值即可

[Code]

1:       int maxPoints(vector<Point> &points) {       
2:            unordered_map<float, int> statistic;   
3:            int maxNum = 0;       
4:            for (int i = 0; i< points.size(); i++)       
5:            {         
6:                 statistic.clear();   
7:                 statistic[INT_MIN] = 0; // for processing duplicate point  
8:                 int duplicate = 1;        
9:                 for (int j = 0; j<points.size(); j++)        
10:                 {         
11:                      if (j == i) continue;          
12:                      if (points[j].x == points[i].x && points[j].y == points[i].y) // count duplicate  
13:                      {            
14:                           duplicate++;            
15:                           continue;          
16:                      }          
17:                      float key = (points[j].x - points[i].x) == 0 ? INT_MAX :   
18:                                     (float) (points[j].y - points[i].y) / (points[j].x - points[i].x);       
19:                      statistic[key]++;         
20:                 }   
21:                 for (unordered_map<float, int>::iterator it = statistic.begin(); it != statistic.end(); ++it)       
22:                 {          
23:                      if (it->second + duplicate >maxNum)          
24:                      {           
25:                           maxNum = it->second + duplicate;          
26:                      }      
27:                 }      
28:            }   
29:            return maxNum;  
30:       }  

若干注意事项：

1. 垂直曲线， 即斜率无穷大

2. 重复节点。