[CareerCup] 7.6 The Line Passes the Most Number of Points 經過最多點的直線

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

 

這道題給了我們許多點，讓我們求經過最多點的一條直線。給之前那道7.5 A Line Cut Two Squares in Half 平均分割兩個正方形的直線一樣，都需要自己寫出點類和直線類。在直線類中，我用我們用斜率和截距來表示直線，為了應對斜率不存在情況，我們還需用一個flag來標記是否為垂直的線。在直線類中，我們要有判斷兩條直線是否相等的函數。判斷相等的方法和之前那道7.3 Line Intersection 直線相交相同，都需要使用epsilon，只要兩個數的差值的絕對值小於epsilon，我們就認定是相等的。對於給定的所有點，每兩個點能組成一條直線，我們的方法是遍曆所有的直線，把所有相同的直線都存入雜湊表中，key是直線的斜率，映射關係是斜率和直線集合的映射，那麼我們只需找到包含直線最多的那個集合即可，參見代碼如下：

 

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<Point> &points) {        Line res(points[0], points[1]);        int bestCnt = 0;        unordered_map<double, vector<Line> > 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<double, vector<Line> > &m, Line &line) {        vector<Line> 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<double, vector<Line> > &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<Line> &lines, Line &line) {        if (lines.empty()) return 0;        int res = 0;        for (auto &a : lines) {            if (a.isEquivalent(line)) ++res;        }        return res;    }};

 

[CareerCup] 7.6 The Line Passes the Most Number of Points 經過最多點的直線