Leedcode Max Points on a Line

1.Description

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

    input parameter: vector<Point> &points

    output parameter: int 

2.Analysis

     A. How to define a line on a 2D plane?

         y = kx + b   y -y0 = k (x - x0) 

        Note: These equations can't describe line without slope.

     B. we are supplied with list of points.

         with points how to get k and b?

         A line is unique with two different points on its given.

         when slope exists, k = (y1 - y0)/(x1 - x0)

         when slope not exists, k = INT_MAX

         So, we can scan every points, using equation y -y0 = k (x - x0) to represent one line. 

                use hash table map to store the line numbers setting k as key and points on line as value.

    C. duplicate

         Every line need add the duplicate num

3.Code

    if (points.size() <= 2)    {        return points.size();    }            std::map<double, int> lines;    std::map<double, int>::iterator it;    double k = 0; //for slope    int i = 0; //for loop    int j = 0; // for loop    int max_points = 0;    int size = points.size();    int x0 = 0;    int y0 = 0;    int x1 = 0;    int y1 = 0;    int duplicate = 0;        for (i = 0; i < size; ++i)    {        x0 = points[i].x;        y0 = points[i].y;        //set variable default        lines.clear();        duplicate = 0;        for (j = 0; j < size; ++j)        {            x1 = points[j].x;            y1 = points[j].y;            // duplicate             if (x1 == x0 && y1 == y0)            {                duplicate++;            }            else            {                // test whether the k exists or not                 k = x1 == x0 ? INT_MAX: (1.0*(y1 - y0))/(x1 - x0);                // line[k] set default 0                lines[k]++;            }        }        // aimed at reduce the computation in compare procedure        max_points -= duplicate;        for (it = lines.begin(); it != lines.end(); ++it)        {            max_points = max_points >= (it->second) ? max_points:(it->second);        }        max_points += duplicate;         //if the lines is empty      very important!!!        if (lines.empty())        {            max_points = duplicate;        }    }    return max_points;

4.Enhancement

We need a*n^2 computation using the above algorithm.  The computation procedure is shown in below table.

 p1p2p3p4p1D k12 k13 k14p2k12Dk23k24p3k13k23Dk34p4k14k24k34D

It is obviously that there exists some duplicated computation.

e.g. the max_point line has point p2 , p3 ,p4 with slope k

then k23 = k24 = k34

we can remove the duplicate computation, then the computation will be reduced more then half.

 duplicate = 1;        for (j = i - 1; j >= 0; --j)        {            x1 = points[j].x;            y1 = points[j].y;            // duplicate             if (x1 == x0 && y1 == y0)            {                duplicate++;            }            else            {                // test whether the k exists or not                 k = x1 == x0 ? INT_MAX: (1.0*(y1 - y0))/(x1 - x0);                // line[k] set default 0                lines[k]++;            }        }

NOTE:

1. For one problem, analyse is very important?

    how to define

    extreme situation

    --> test scenarios

2.coding

    A proper framework and data struct can benefits lot.