Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
分析: 任取两点组成的直线,看有多少点在其上面。
注意: 考虑 3 种情况: 两点重合; 斜率为无穷大; 正常情况
/**
* Definition for a point.
* struct Point {
* int x;
* int y;
* Point() : x(0), y(0) {}
* Point(int a, int b) : x(a), y(b) {}
* };
*/
class Solution {
public:
int maxPoints(vector<Point> &points) {
int max_points(0);
double a(0), b(0); // denote a straight line
LINE_TYPE line_type(LINE);
if(points.size() == 0) return 0;
else if(points.size() == 1) return 1;
for(int i=0; i<points.size(); ++i)
for(int j=i+1; j<points.size(); ++j)
{
if(points[i].x == points[j].x){
if(points[i].y == points[j].y)
line_type = POINT;
else
line_type = VLINE;
}
else{
line_type = LINE;
a = (points[j].y - points[i].y) / static_cast<double>(points[j].x - points[i].x); //tangent
b = points[i].y - a * points[i].x;
}
// compute how many points on the line
int point_count(2);
for(int k=0; k<points.size(); ++k){
if(k == i || k == j) continue;
// test whether points[k] on the line
if(line_type == LINE){
if( fabs(points[k].y - a*points[k].x - b) < 1e-10)
++ point_count;
}
else if(line_type == POINT){
if(points[k].x == points[i].x && points[k].y == points[i].y)
++ point_count;
}
else{
if(points[k].x == points[i].x)
++ point_count;
}
}
max_points = max(max_points, point_count);
}
return max_points;
}
private:
enum LINE_TYPE{POINT, LINE, VLINE};
};LeetCode----Max Points On a Line
原文:http://blog.csdn.net/shoulinjun/article/details/19152791