二维坐标中在一条直线上最大点数

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import java.util.HashMap;
class Point {int x;int y;Point() { x = 0; y = 0; }Point(int a, int b) { x = a; y = b; }}public class Solution {//解法一:存在问题public int maxPoints(Point[] points) {if(points==null||points.length==0)return 0;if(points.length==1||points.length==2)return points.length;HashMap<Integer,Integer>map=new HashMap<>();  for(int i=0;i!=points.length;i++){for(int j=i+1;j!=points.length;j++){int k=getXie(points[i],points[j]);if(!map.containsKey(k)){map.put(k,1);}else{int temp=map.get(k);map.put(k,++temp);}}}int max=0;for(int k:map.keySet()){max=Math.max(max,map.get(k));}return max;}//获得两个点的斜率public int getXie(Point point1,Point point2){int a= Math.abs(point2.y-point1.y);int b=Math.abs(point2.x-point1.x);if(a==0)return 0;if(b==0)return Integer.MAX_VALUE;return  a/b;}//方法二:斜率求解法public int maxPoints2(Point[] points) {if(points==null||points.length==0)return 0;if(points.length==1||points.length==2)return points.length;int max=0;for(int i=0;i!=points.length;i++){int vcnt=0;  //垂直的点int dup=0;   //重复的点int curmax=1;HashMap<Double,Integer>map=new HashMap<>();for(int j=0;j!=points.length;j++){if(j!=i){double x1=points[i].x-points[j].x;double y1=points[i].y-points[j].y;if(x1==0&&y1==0){dup++;   //重复点加1}else if(x1==0){ if(vcnt==0)vcnt=2;elsevcnt++;  //垂直点加1curmax=Math.max(vcnt,curmax);}else{double k=y1/x1;  //斜率if(!map.containsKey(k)){map.put(k,2); }else{int temp=map.get(k);map.put(k,++temp);    }curmax=Math.max(map.get(k),curmax);}}}max=Math.max(max,curmax+dup);}return max;}public static void main(String[]args){//System.out.println("Hello World!");Point point1=new Point(1,1);Point point2=new Point(2,2);Point point3=new Point(3,3);Point point4=new Point(3,4);Point[] points={point1,point2,point3,point4};Solution s=new Solution();System.out.println(s.maxPoints2(points));}
}