• [CareerCup] 7.6 The Line Passes the Most Number of Points 经过最多点的直线


    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7.5 A Line Cut Two Squares in Half 平均分割两个正方形的直线一样,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7.3 Line Intersection 直线相交相同,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<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;
        }
    };
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  • 原文地址:https://www.cnblogs.com/grandyang/p/4779741.html
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