[LeetCode]Max Points on a Line

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.

对点遍历，找到对每个点共线的最大点数，所有点的最大值即是全局最大值。

Note：对每个值用Hash table记录当前点和其他点的斜率，key为斜率值，竖直的线的斜率位double(INT_MAX)，value是出现的个数；

         可能出现相同坐标的点。

/**
 * 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) {
        if(points.size()<3) return points.size();
        int result=0;
        unordered_map<double,int> showed;
        for(int i=0;i<points.size()-1;i++)
        {
            showed.clear();
            int samePoint=1;
            for(int j=i+1;j<points.size();j++)
            {
                if(points[i].x==points[j].x)
                {
                    if(points[i].y==points[j].y)
                    {
                        samePoint++;
                    }
                    else
                    {
                        if(showed.find(double(INT_MAX))!=showed.end())
                        {
                            showed[double(INT_MAX)]+=1;
                        }
                        else
                        {
                            showed[double(INT_MAX)]=1;
                        }
                    }
                }
                else
                {
                    double slope=(points[j].y-points[i].y)*1.0/(points[j].x-points[i].x);
                    if(showed.find(slope)!=showed.end())
                    {
                        showed[slope]+=1;
                    }
                    else
                    {
                        showed[slope]=1;
                    }
                }
            }
            int max_i=0;
            for(unordered_map<double,int>::iterator iter=showed.begin();iter!=showed.end();iter++)
            {
                if(max_i<iter->second)
                {
                    max_i=iter->second;
                }
            }
            max_i=max_i+samePoint;
            if(result<max_i)
            {
                result=max_i;
            }
        }
        
        return result;
    }
};