Max Points on a Line

最新推荐文章于 2018-07-30 23:28:01 发布

原创最新推荐文章于 2018-07-30 23:28:01 发布 · 413 阅读

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本文介绍了一个算法，用于确定二维平面上给定点集中，有多少点位于同一直线上，并详细解释了函数接口、算法思想及实现过程。

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

函数接口：int maxPoints(vector<Point> &points) {}

算法思想：先剔除点数组points中重复的点，然后从前两个点开始，构建一条直线，依次加入新的点，新加入的点要么加到原有的直线上，要么形成新的直线，直至所有点加入。然后统计所有直线上的点数（加上重复的点），得到最大值。

int maxPoints(vector &points) 
    {
      	if (points.size() == 0)
    		return 0;
    
        vector points_nodup;
        unordered_map mark;
    	stringstream ss;
    	string s;
         
    	for (int i = 0; i < points.size(); i++)
    	{
    		ss.str("");
    		ss << points[i].x << '-' << points[i].y;
    		s = ss.str();
    
    		if (mark.find(s) == mark.end())
    		{
    			points_nodup.push_back(points[i]);
    			mark[s] = 1;
    		}
    		else
    			mark[s]++;		
    	}
            
        if (points_nodup.size() == 1)
        	return points.size();
        		
        if (points_nodup.size() == 2)
        	return points.size();
        	
        vector> lines;
        vector line;
        
        line.push_back(0);
        line.push_back(1);
        lines.push_back(line);
        
        for (int i = 2; i < points_nodup.size(); i++)
        {
        	Point p = points_nodup[i];
        	bool flag = false;
        
        	for (int j = 0; j < lines.size(); j++)
        	{
        		if ((points_nodup[lines[j][1]].y-points_nodup[lines[j][0]].y)*(p.x-points_nodup[lines[j][1]].x) ==
        			(p.y-points_nodup[lines[j][1]].y)*(points_nodup[lines[j][1]].x-points_nodup[lines[j][0]].x))
        		{
        			lines[j].push_back(i);
        			flag = true;
        			break;
        		}
        	}
        
        	if (flag == false)
        	{
        		for (int j = 0; j < i; j++)
        		{
        			line.clear();
        			line.push_back(j);
        			line.push_back(i);
        			lines.push_back(line);
        		}
        	}
        }
        
        int maxNum = 0;
        
        for (int i = 0; i < lines.size(); i++)
        {
    		int pointNum = 0;
    
    		for (int j = 0; j < lines[i].size(); j++)
    		{
    			ss.str("");
    			ss << points_nodup[lines[i][j]].x << '-' << points_nodup[lines[i][j]].y;
    			s = ss.str();
    
    			pointNum += mark[s];
    		}
    
    		if (maxNum < pointNum)
    			maxNum = pointNum;	
        }
        	
        return maxNum;
    }