[LeetCode] Max Points on a Line

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

Solution:

/**
* 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:
double PI = 3.1415926;

double computeAngle(int x, int y, int xx, int yy)
{
if(x == xx)
{
if(y == yy)
return 2 * PI;
else
return PI;
}
else
return atan((double)(yy - y) / (xx - x));
}

int maxPoints(vector<Point> &points) {
int n = points.size();
if(n <= 2) return n;
double **angle = new double*[n + 1];
int **visited = new int*[n + 1];
for(int i = 0;i < n;i++)
{
angle[i] = new double[n + 1];
visited[i] = new int[n + 1];
}
for(int i = 0;i < n;i++)
{
memset(visited[i], 0, (n + 1) * sizeof(int));
for(int j = i;j < n;j++)
angle[i][j] = angle[j][i] = computeAngle(points[i].x, points[i].y, points[j].x, points[j].y);
}
int max = 2;
for(int i = 0;i < n;i++)
{
for(int j = 0;j < n;j++)
{
if(i == j || visited[i][j] == 1) continue;
visited[i][j] = 1;
//define a line by point i and j
double curAngle = angle[i][j];
int curNum = 2;
for(int k = 0;k < n;k++)
{
if(i != k && j != k)
{
if(curAngle == 2 * PI)
{
if(angle[i][k] == 2 * PI || angle[j][k] == 2 * PI)
{
curNum++;
}
else
{
curNum++;
curAngle = angle[i][k];
visited[i][k] = visited[j][k] = visited[k][j] = visited[k][i] = 1;
}
}
else
{
if(angle[i][k] == curAngle)
{
curNum++;
visited[i][k] = visited[j][k] = visited[k][j] = visited[k][i] = 1;
}
}
}
}
//  cout << i << " " << j << " :" <<curNum << endl;
max = (curNum > max) ? curNum : max;
}
}
return max;
}
};