题目

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

题解

这道题就是给你一个2D平面,然后给你的数据结构是由横纵坐标表示的点,然后看哪条直线上的点最多。

(1)两点确定一条直线

(2)斜率相同的点落在一条直线上

(3)坐标相同的两个不同的点 算作2个点

利用HashMap,Key值存斜率,Value存此斜率下的点的个数。同时考虑特殊情况,如果恰巧遍历到一个相同坐标的点,那么就维护一个local的counter来记录相同的点。

维护一个localmax,计算当前情况下的最大值;再维护一个全局Max来计算总的最大值。

返回全局Max即可。

代码如下:

 1     public int maxPoints(Point[] points) {  

 2         if(points.length == 0||points == null) 

 3             return 0;  

 4             

 5         if(points.length == 1) 

 6             return 1;  

 7             

 8         int max = 1;  //the final max value, at least one

 9         for(int i = 0; i < points.length; i++) {  

             HashMap<Float, Integer> hm = new HashMap<Float, Integer>();  

             int same = 0;

             int localmax = 1; //the max value of current slope, at least one

             for(int j = 0; j < points.length; j++) {  

                 if(i == j) 

                     continue;  

                     

                 if(points[i].x == points[j].x && points[i].y == points[j].y){

                     same++; 

                     continue;

                 }

                 

                 float slope = ((float)(points[i].y - points[j].y))/(points[i].x - points[j].x); 

                 

                 if(hm.containsKey(slope))  

                     hm.put(slope, hm.get(slope) + 1);  

                 else  

                     hm.put(slope, 2);  //two points form a line

             }

             

             for (Integer value : hm.values())   

                 localmax = Math.max(localmax, value);  

           

             localmax += same;  

             max = Math.max(max, localmax);  

         }  

         return max; 

     }

Reference:

http://blog.csdn.net/ttgump/article/details/23146357

http://blog.csdn.net/linhuanmars/article/details/21060933

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