LeetCode----Max Points On a Line

Max Points on a Line

Total Accepted: 4246
Total Submissions: 42602

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};
};

再给出一个 N^2的代码：

int maxPoints(vector<Point> &points) {
if(points.size() < 2) return points.size();
int res(1), sameCount(0), verticalCount(0);
unordered_map<double, int> tangent;

for(int i=0; i<points.size(); ++i)
{
sameCount = 1;
verticalCount = 0;
tangent.clear();
for(int j=0; j<points.size(); ++j)
{
if(j == i) continue;
if(points[j].x == points[i].x) {
if(points[j].y == points[i].y) ++ sameCount;
else ++ verticalCount;
}
else {
double k = (points[j].y-points[i].y) / double(points[j].x-points[i].x);
++ tangent[k];
}
}
for(auto x : tangent) {
res = max(res, x.second+sameCount);
}
res = max(res, verticalCount+sameCount);
}
return res;
}