[LeetCode]Max Points on a Line

 这题的主要思路是建立一个二维的数组,然后遍历该二维数组,其实一个上三角或者下三角的矩阵也就足够了,然后注意把已经搜索过的解给标记,避免重复搜索,最有有一点要注意的是浮点数double的判断相等不能直接用等于,因为会有舍入误差,因此设计了一个判断相等的函数,只要两个浮点数相差小于一个很小的小数,就认为它们相等。

class Solution {
public:
	int maxPoints(vector<Point> &points) {
		int size = points.size();
		if (points.size() < 2)
			return points.size();
		bool **a = new bool*[size];
		for (int i = 0; i < size; i++)
		{
			a[i] = new bool[size];
			for (int j = 0; j < size; j++)
			{
				a[i][j] = false;
			}
		}
		int max = 2;
		int count = 2;
		for (int i = 0; i < size;i++)
			for (int j = 0; j < size; j++)
		{
				if (i == j)continue;
				if (!a[i][j])
				{
					a[i][j] = a[j][i] = true;
					if (points[i].x == points[j].x)
						vertical = true;
					else vertical = false;

					SetXY(points[i], points[j]);
					//cout << "line:(" << points[i].x << ","<<points[i].y << ")(" << points[j].x << ","<<points[j].y << ")" << endl;
					count = 2;
					for (int k = 0; k < size; k++)
					{
						if (k != i&&k != j&&OnSameLine(points[k]))
						{
							//cout<<
							a[i][k] = a[j][k] = a[k][i] = a[k][j] = true;
							count++;
						}
					}
					if (count>max)
						max = count;
				}				
		}
		for (int i = 0; i < size;i++)
		{
		    delete []a[i];
		}
		delete []a;
			return max;
	}
private:
	Point x;
	Point y;
	double a;
	bool vertical;
	void SetXY(Point& x, Point&y)
	{
		this->x = x;
		this->y = y;
		if (!vertical)
			a = (double)(this->x.y - this->y.y) / (double)(this->x.x - this->y.x);	
	}

	bool OnSameLine(Point& z)
	{
		if (!vertical)
		{
			double temp = (z.y - y.y) - a * (z.x - y.x);
			if (temp < 0)
				temp = -temp;
			return temp < 0.000001;
		}
		else return z.x == y.x;
	}

};


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