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; } };
本文转自博客园Grandyang的博客，原文链接：经过最多点的直线[CareerCup] 7.6 The Line Passes the Most Number of Points，如需转载请自行联系原博主。
