Max Points on a Line--LeetCode

1.题目

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.

2.题意

计算几何，给定一个点集合，求最大的共线点的个数

3.分析

找出相同点和相同斜率的点

通过斜率来判断共线需要用到除法
用double表示的双精度小数在有的系统里不一定准确
为了精确无误的计算，应当避免除法，避免无线不循环小数的出现
可将两数分别除以它们的最大公约数，pair作为key
gcd采用算法导论中的欧几里德算法

注意两个max函数要放在外层循环，不要忘记return res;

4.代码

/** * 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) {        int res = 0;        for(int i = 0; i < points.size(); ++i)        {            int duplicate = 1;            map<pair<int, int>, int> m;            for(int j = i + 1; j < points.size(); ++j)            {                if(points[j].y == points[i].y && points[j].x == points[i].x)                    ++duplicate;                else                {                    int dy = points[j].y - points[i].y;                    int dx = points[j].x - points[i].x;                    int d = gcd(dx, dy);                    ++m[{dx/d, dy/d}];                }            }            res = max(res, duplicate);            for(auto it = m.begin(); it != m.end(); ++it)                res = max(res, it->second + duplicate);        }        return res;    }private:    int gcd(int a, int b)    {        if(b == 0)            return a;        else            return gcd(b, a % b);    }};