Max Point on a line

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

1.N个顶点在2D plane ,找到其中点最多的那条直线

 1）使用斜率判断；注意K=INF 两个点是在同一条输入的垂直线上

/**
 * Definition for a point.
 * struct Point {
 *     int x;
 *     int y;
 *     Point() : x(0), y(0) {}
 *     Point(int a, int b) : x(a), y(b) {}
 * };
 */
class Solution {
public:
    int maxPoints(vector<Point> &points) {
        
        /*找到一条直线上最大数目的点,返回这条直线*/
        if(points.size()<3)
            return points.size();
        
       
        int maxNum=0; map<double,int> slope_count;
        /*以点为中心，计算其他n-1点与之斜率，之中挑选出最大点的斜率点*/
        for(int i=0;i<points.size();i++)
        
         {
            int cnt=0;//b表示重合点数
             slope_count.clear();
            for(int j=i+1;j<points.size();j++)
             {
                
                  if(i==j) continue;
                
                /*点是不是重合点*/
                
                if(points[i].x == points[j].x && points[i].y == points[j].y)
                {
                    cnt++;
                    continue;
                }  
                
                /*判断斜率，注意与Y轴平行的*/
                
                double slope=points[i].x==points[j].x ? INT_MAX:(double)(points[j].y - points[i].y)/(points[j].x - points[i].x); 
                
                slope_count[slope]++;
            }
            
        
        
        /*对其进行比较*/
         if(slope_count.size()==0)
               slope_count[INT_MIN]=0;/*所有点与Y轴平行*/
        
         map<double,int>::iterator it = slope_count.begin();
         
         for(;it!=slope_count.end();it++)
              {
                        if(it->second+cnt>maxNum)
                            maxNum=it->second+cnt;
            
               }
        }
        /*最开始并没有记录其开始点，所有最后+1*/
        return maxNum+1;
    }
};