【LeetCode】149. 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.

点和方向确定一条直线。

需要两重循环，第一重循环遍历起始点a，第二重循环遍历剩余点b。

a和b如果不重合，就可以确定一条直线。

对于每个点a，构建 斜率->点数 的map。

(1)b与a重合，以a起始的所有直线点数+1 (用dup统一相加)

(2)b与a不重合，a与b确定的直线点数+1

/**
 * 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.empty())
            return ;
        else if(points.size() == )
            return ;

        int ret = ;
        for(int i = ; i < points.size(); i ++)
        {//start point
            int curmax = ; //points[i] itself
            unordered_map<double, int> kcnt;    // slope_k count
            int vcnt = ;   // vertical count
            int dup = ;    // duplicate added to curmax
            for(int j = ; j < points.size(); j ++)
            {
                if(j != i)
                {
                    double deltax = points[i].x - points[j].x;
                    double deltay = points[i].y - points[j].y;
                    if(deltax ==  && deltay == )
                        dup ++;
                    else if(deltax == )
                    {
                        if(vcnt == )
                            vcnt = ;
                        else
                            vcnt ++;
                        curmax = max(curmax, vcnt);
                    }
                    else
                    {
                        double k = deltay / deltax;
                        if(kcnt[k] == )
                            kcnt[k] = ;
                        else
                            kcnt[k] ++;
                        curmax = max(curmax, kcnt[k]);
                    }
                }
            }
            ret = max(ret, curmax + dup);
        }
        return ret;
    }
};