[LeetCode OJ] 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.

//定义二维平面上的点
struct Point
{
    int x;
    int y;
    Point(int a=0, int b=0):x(a),y(b){}
};

bool operator==(const Point& left, const Point& right)
{
    return (left.x==right.x && left.y==right.y);
}

//求两个点连接成的直线所对应的斜率
double line_equation(const Point& P1, const Point& P2)
{
    double slope;
    if(P1.x==P2.x)
        slope = 1e+6;
    else
        slope = double(P1.y-P2.y)/(P1.x-P2.x);
    return slope;
}

class Solution {
public:
    int maxPoints(vector<Point> &points) 
    {
        int max=0;
        for(vector<Point>::iterator it=points.begin(); it!=points.end(); ++it)
        {
            map<double, int> Line_count;
            int same_count=0;    //same_count表示同一个点重复出现的个数
            
            for(vector<Point>::iterator it2=it; it2!=points.end(); ++it2)   //此处it2=points.begin()还是it2=it虽然不影响最终结果，但是对于时间复杂度影响很大，it2=it的时间复杂度要低很多
            {
                if(*it==*it2)
                {
                    ++same_count;
                    continue;
                }    
                double slope = line_equation(*it, *it2);
                ++Line_count[slope];
            }
            set<int> iset;
            for(map<double, int>::iterator iter = Line_count.begin(); iter!=Line_count.end(); ++iter)
                iset.insert(iter->second);
            int max_now = same_count;
            if(iset.begin()!=iset.end())
                max_now = same_count + *(--iset.end());
            if(max_now>max)
                max = max_now;
        }
        return max;
    }
};