leetcode: Max Points on a Line 题解

题目：

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

 

思路：

 （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

使用Hashmap来存放键值对，Key用来存放斜率，value存放该斜率下的点的个数，对每个点进行循环遍历，计算该点和其余各个点的斜率，然后更新Hashmap中对应的值，每个点都计算出点数最大的值，保存在局部变量localmax中，维护一个全局变量max来存放最后的结果。

 

public int maxPoints(Point[] points) {  
        if(points.length == 0||points == null) 
            return 0;  
            
        if(points.length == 1) 
            return 1;  
            
        int max = 1;  //the final max value, at least one
        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; 
    }