[leedcode 149] Max Points on a 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) {
        //三层遍历，依次判断是否这些点在一条线路。注意判断两个相同的点的情况。
        //相同点的个数使用一个计数器dup，非相同但在一条直线上也有一个变量cur
        if(points==null||points.length<=0) return 0;
        int len=points.length;
        int dup=1;
        int cur=1;
        int res=0;
        for(int i=0;i<len;i++){
            dup=1;
            for(int j=i+1;j<len;j++){
                if(points[i].x==points[j].x&&points[i].y==points[j].y){//相同点
                    dup++;
                    continue;
                }
                for(int k=j+1;k<len;k++){
                    if(isSameLine(points[i],points[j],points[k])){
                        cur++;
                    }
                    
                }
                if(res<cur+dup)res=cur+dup;
                cur=1;
            }
            if(res<dup)res=dup;//注意都是相同点的处理
            cur=1;
        }
        return res;
    }
    public boolean isSameLine(Point i,Point j,Point k){//相同点的判断，斜率相同
        if((i.x-j.x)*(i.y-k.y)==(i.x-k.x)*(i.y-j.y)) return true;
        else return false;
        
    }
}