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.

思路：这道题主要是找出重叠结点和相同斜率之和。此时我们要判断有斜率和无斜率个数，并判断那个数量比较大，更新斜率最大值。然后在与相同情况下重叠结点个数累加。最后就是更新同一条直线上点数最大值。

/**

 * 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() <= 2){

            return points.size();

        }

        

        map<double,int> slop_lines;

        int maxNumber = 0, same_points, none_slop;

        double slop;

        for(int i = 0; i < points.size()-1; ++i){

            slop_lines.clear();

            same_points = 1;

            none_slop = 0;

            for(int j = i+1; j < points.size(); ++j){

                if(points[i].x == points[j].x &&  points[i].y == points[j].y)

                {

                    ++same_points;

                    continue;

                }

                if(points[i].x == points[j].x){

                    ++none_slop;

                }else{

                    slop = (double)(points[i].y - points[j].y)/(points[i].x - points[j].x);

                    if(slop_lines.find(slop) != slop_lines.end()){

                        ++slop_lines[slop];

                    }else{

                        slop_lines[slop] = 1;

                    }

                }

            }

            

            //更新最大值

            map<double,int>::iterator iter = slop_lines.begin();

            for(;iter != slop_lines.end(); ++iter){

                if(iter->second > none_slop){

                    none_slop = iter->second;

                }

            }

            none_slop += same_points;

            if(none_slop > maxNumber){

                maxNumber=none_slop;

            }

        }

        

        return maxNumber;

    }

};