[CareerCup] 7.6 The Line Passes the Most Number of Points 经过最多点的直线 - Grandyang - 博客园
Given a two-dimensional graph with points on it, find a line which passes the most number of points.
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Read full article from [CareerCup] 7.6 The Line Passes the Most Number of Points 经过最多点的直线 - Grandyang - 博客园
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，
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; } };
We will represent a line as a slope and y-intercept (as opposed to a pair of points), which allows us to easily check to see if the line from ( xl, yl) to ( x2, y2) is equivalent to the line from ( x3, y3) to ( x4,
y4).
floating point numbers cannot always be represented accurately in binary. We resolve this by checking if two floating point numbers are within an epsilon value of each other
It means that two lines with "equal" slopes may not be hashed to the same value. To solve this, we will round the slope down to the next epsilon and use this flooredSlope as the hash key. Then, to retrieve all lines that are potentially equal, we will search the hash table at three spots: flooredSlope, flooredSlope - epsilon, and flooredSlope + epsilon. This will ensure
that we've checked out all lines that might be equal.
/* Find line that goes through most number of points. */
Line findBestLine(GraphPoint[] points) {
HashMapList<Double, Line> linesBySlope = getListOfLines(points);
return getBestLine(linesBySlope);
}
/* Add each pair of points as a line to the list. */
HashMapList<Double, Line> getListOfLines(GraphPoint[] points) {
HashMapList<Double, Line> linesBySlope = new HashMapList<Double, Line>();
for (int i= 0; i < points.length; i++) {
for (int j = i + 1; j < points.length; j++) {
Line line= new Line(points[i], points[j]);
double key= Line.floorToNearestEpsilon(line.slope);
linesBySlope.put(key, line);
}
}
return linesBySlope;
}
/* Return the line with the most equivalent other lines. */
Line getBestLine(HashMapList<Double, Line> linesBySlope) {
Line bestLine null;
int bestCount = 0;
Set<Double> slopes = linesBySlope.keySet();
for (double slope : slopes) {
ArrayList<Line> lines = linesBySlope.get(slope);
for (Line line : lines) {
// count lines that are equivalent to current line
int count = countEquivalentLines(linesBySlope, line);
// if better than current line, replace it
if (count> bestCount) {
bestLine = line;
bestCount = count;
bestLine.Print();
System.out.println(bestCount);
}
}
}
return bestLine;
}
/* Chec k hashmap for lines that are equivalent. Note that we need to check one
* epsilon above and below the actual slope since we're defining two lines as
* equivalent if they're within an epsilon of each other. */
int countEquivalentLines(HashMapList<Double, Line> linesBySlope, Line line) {
double key= Line.floorToNearestEpsilon(line.slope);
int count = countEquivalentLines(linesBySlope.get(key), line);
count+= countEquivalentLines(linesBySlope.get(key - Line.epsilon), line);
count += countEquivalentLines(linesBySlope.get(key + Line.epsilon), line);
return count;
}
/* Count lines within an array of lines which are "equivalent" (slope and
* y-intercept are within an epsilon value) to a given line */
int countEquivalentLines(ArrayList<Line> lines, Line line) {
if (lines == null) return 0;
int count = 0;
for (Line parallelLine : lines) {
if (parallelLine.isEquivalent(line)) {
count++;
}
}
return count;
}
public class Line {
public static double epsilon= .0001;
public double slope, intercept;
private boolean infinite_slope = false;
public Line(GraphPoint p, GraphPoint q) {
if (Math.abs(p.x - q.x) > epsilon) { // if x's are different
slope= (p.y - q.y)I (p.x - q.x); II compute slope
intercept= p.y - slope * p.x; II y intercept from y=mx+b
} else {
infinite_slope = true;
intercept = p.x; II x-intercept, since slope is infinite
}
}
public static double floorToNearestEpsilon(double d) {
int r = (int) (d / epsilon);
return ((double) r) * epsilon;
}
public boolean isEquivalent(double a, double b) {
return (Math.abs(a - b) < epsilon);
}
public boolean isEquivalent(Object o) {
Line 1 = (Line) o;
if (isEquivalent(l.slope, slope) && isEquivalent(l.intercept, intercept) &&
(infinite_slope == l.infinite_slope)) {
return true;
}
return false;
}
}
Read full article from [CareerCup] 7.6 The Line Passes the Most Number of Points 经过最多点的直线 - Grandyang - 博客园
