7.6 Given a two-dimensional graph with points on it, find a line which passes the most number of points.

 

这道题给了我们许多点，让我们求经过最多点的一条直线。给之前那道一样，都需要自己写出点类和直线类。在直线类中，我用我们用斜率和截距来表示直线，为了应对斜率不存在情况，我们还需用一个flag来标记是否为垂直的线。在直线类中，我们要有判断两条直线是否相等的函数。判断相等的方法和之前那道相同，都需要使用epsilon，只要两个数的差值的绝对值小于epsilon，我们就认定是相等的。对于给定的所有点，每两个点能组成一条直线，我们的方法是遍历所有的直线，把所有相同的直线都存入哈希表中，key是直线的斜率，映射关系是斜率和直线集合的映射，那么我们只需找到包含直线最多的那个集合即可，参见代码如下：

 

class Point {public:    double _x, _y;    Point(double x, double y): _x(x), _y(y) {};};class Line {public:    static constexpr double _epsilon = 0.0001;    double _slope, _intercept;    bool _infi_slope = false;    Line(Point p, Point q) {        if (fabs(p._x - q._x) > _epsilon) {            _slope = (p._y - q._y) / (p._x - q._x);            _intercept = p._y - _slope * p._x;        } else {            _infi_slope = true;            _intercept = p._x;        }    }    static double floorToNearestEpsilon(double d) {        int r = (int)(d / _epsilon);        return ((double)r) * _epsilon;    }    bool isEquivalent(double a, double b) {        return (fabs(a - b) < _epsilon);    }    bool isEquivalent(Line other) {        if (isEquivalent(_slope, other._slope) && isEquivalent(_intercept, other._intercept) && (_infi_slope == other._infi_slope)) {            return true;        }        return false;    }};class Solution {public:    Line findBestLine(vector
     
       &points) {        Line res(points[0], points[1]);        int bestCnt = 0;        unordered_map
      
        > m;        for (int i = 0; i < (int)points.size(); ++i) {            for (int j = i + 1; j < (int)points.size(); ++j) {                Line line(points[i], points[j]);                insertLine(m, line);                int cnt = countEquivalentLines(m, line);                if (cnt > bestCnt) {                    res = line;                    bestCnt = cnt;                }            }        }        return res;    }    void insertLine(unordered_map
       
         > &m, Line &line) {        vector
        
          lines;        double key = Line::floorToNearestEpsilon(line._slope);        if (m.find(key) != m.end()) {            lines = m[key];        } else {            m[key] = lines;        }        lines.push_back(line);    }    int countEquivalentLines(unordered_map
         
           > &m, Line &line) {        double key = Line::floorToNearestEpsilon(line._slope);        double eps = Line::_epsilon;        return countEquivalentLines(m[key], line) + countEquivalentLines(m[key - eps], line) + countEquivalentLines(m[key + eps], line);    }    int countEquivalentLines(vector
          
            &lines, Line &line) { if (lines.empty()) return 0; int res = 0; for (auto &a : lines) { if (a.isEquivalent(line)) ++res; } return res; }};