max-points-on-a-line java code

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

/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */
public class Solution {
    public int maxPoints(Point[] points) {
        //关键在于判断三点共线，两平行直线有且只有一个交点，所以有一个中间点，这个中间点与另外两个端点的连线的斜率相等
        //由比率的性质
        int ABx;
        int ABy;
        int BCx;
        int BCy;

        if(points.length<=) return points.length;
        int max=;//用来记录最大个数

        for(int i=;i<points.length;i++){
            int num=;
            int temp=;

            for(int j=i+;j<points.length;j++){
                ABx=points[i].x-points[j].x;
                ABy=points[i].y-points[j].y;
                if(ABx== && ABy==)//表示出现重复点
                {
                    num++;
                }else{
                    temp++;
                    for(int k=j+;k<points.length;k++){
                        BCx=points[j].x-points[k].x;
                        BCy=points[j].y-points[k].y;
                        if(ABx*BCy==BCx*ABy){//表示两个斜率相等，转化为乘积的形式可以避免分母为0的情况
                            temp++;
                        }
                    }
                }
                if(max<(num+temp)){
                  max=num+temp;
                }
                temp=;
            }

        }
        return max;
    }
}