leetcode 149 Max Points on a Line

参考： https://www.cnblogs.com/grandyang/p/4579693.html

https://www.iteye.com/blog/yiminghe-568666 三点共线问题判断

解法一：从第一个节点开始，找与它在同一条线上的；如果三点共线，则x2-x0/y2-y0=x1-x0/y1-y0；每一轮循环，能找到大于i的点中，与points[i]共线的所有点。

class Solution {
public:
    int maxPoints(vector<vector<int>>& points) {
        int re=0;
        for(int i=0;i<points.size();++i) {
            map<pair<int,int>,int> m;
            int dup=1;//至少有points[i]这个节点
            for(int j=i+1;j<points.size();++j) {
                if(points[j][0]==points[i][0]&&points[j][1]==points[i][1]) {
                    ++dup;continue;
                }
                int dx=points[j][0]-points[i][0];
                int dy=points[j][1]-points[i][1];
                int d=gcd(dx,dy);
                ++m[{dx/d,dy/d}];
            }
            re=max(re,dup);
            for(auto& t:m) {
                re=max(re,t.second+dup);
            }
        }
        return re;
    }
    int gcd(int a,int b) {
        return b==0?a:gcd(b,a%b);
    }
};

解法二：如果三点共线，则三点形成的图形的面积为0.

class Solution {
public:
    int maxPoints(vector<vector<int>>& points) {
        int re=0;
        for(int i=0;i<points.size();++i) {
            int dup=1;
            for(int j=i+1;j<points.size();++j) {
                int cnt=0;
                long long x1=points[i][0],x2=points[j][0],y1=points[i][1],y2=points[j][1];
                if(x1==x2&&y1==y2) {++dup;continue;}
                for(int k=0;k<points.size();++k) {
                    int x3=points[k][0],y3=points[k][1];
                    if(x1*y2+x2*y3+x3*y1-x1*y3-x2*y1-x3*y2==0) ++cnt;
                }
                re=max(re,cnt);
            }
            re=max(re,dup);
        }
        return re;
    }
};