[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;

}
}