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