package leetcode;

 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 MaxPointOnALine {
public static int maxPoints(Point[] points) { if(points.length<=2) {
return points.length;
}
//斜率
double k = 0.0;
int maxPointNum = 0;
int parallelNum =0;
int tempMaxPointNum = 0;
int sameNum=0;
HashMap map=new HashMap<Double,Integer>();
for(int i=0;i<points.length;i++){
sameNum=1;
parallelNum=0;
tempMaxPointNum=0;
map.clear();
for(int j=i+1;j<points.length;j++){
if(points[i].x==points[j].x&&points[i].y==points[j].y){
sameNum++;
}
else if(points[i].x==points[j].x){
parallelNum++; }else {
if(points[i].y==points[j].y){
k=0.0; }else{
k=((double)(points[i].y-points[j].y))/(points[i].x-points[j].x);
}
if(map.get(k)==null){
map.put(k, new Integer(1));
if(1>tempMaxPointNum){
tempMaxPointNum=1;
}
}else{
int number=(int)map.get(k);
number++;
map.put(k, number);
if(number>tempMaxPointNum){
tempMaxPointNum=number;
}
}
}
}
if(parallelNum>1){
if(parallelNum>tempMaxPointNum)
tempMaxPointNum=parallelNum;
}
tempMaxPointNum+=sameNum;
if(tempMaxPointNum>maxPointNum)
maxPointNum=tempMaxPointNum;
}
return maxPointNum;
}
public static void main(String[] args){
Point[] parr = {new Point(1,1),new Point(1,2)};
//maxPoints(parr);
System.out.println(maxPoints(parr));
}
}

leetcode--001 max point on a line的更多相关文章

  1. 【leetcode】Max Points on a Line

    Max Points on a Line 题目描述: Given n points on a 2D plane, find the maximum number of points that lie ...

  2. [LeetCode OJ] Max Points on a Line

    Max Points on a Line Submission Details 27 / 27 test cases passed. Status: Accepted Runtime: 472 ms ...

  3. [LeetCode] 149. Max Points on a Line 共线点个数

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

  4. 【leetcode】Max Points on a Line(hard)☆

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

  5. Java for LeetCode 149 Max Points on a Line

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

  6. leetcode 149. Max Points on a Line --------- java

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

  7. [leetcode]149. Max Points on a Line多点共线

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

  8. [LeetCode OJ] Max Points on a Line—Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

    //定义二维平面上的点struct Point { int x; int y; Point(, ):x(a),y(b){} }; bool operator==(const Point& le ...

  9. LeetCode之Max Points on a Line Total

    1.问题描述 Given n points on a 2D plane, find the maximum number of points that lie on the same straight ...

  10. leetcode[149]Max Points on a Line

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

随机推荐

  1. Lucene中的合并因子mergeFactor

    mergeFactor 是用来决定segment该如何被addDocument()方法进行合并的. 当mergeFactor取比较小的值时,索引时所使用的内存较少 而且搜素未优化的速度会比较快.因此, ...

  2. java 多线程机制

    Example12_1.java public class Example12_1 { public static void main(String args[]) { //主线程 SpeakElep ...

  3. 当使用System,out.println()打印一个对象是自动调用toString方法

    在Java中,所有的对象都是继承自Object,自然继承了toString方法,在当使用System,out.println()里面为一个对象的引用时,自动调用toString方法讲对象打印出来.如果 ...

  4. Node.js学习 - Route

    Node.js 路由 URL解析,需要url和querystring两个模块url.parse(string).query | url.parse(string).pathname | | | | | ...

  5. PAT1064(上)分析部分

    Complete Binary Search Tree (30) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B A Binary Search Tree (BST) ...

  6. 博弈论最简单例子TacTicToe

    博弈论是人工智能中的一个分支.顾名思义就是下棋的算法.当然引申出来的应用可能不止用来下棋,也可以用来做游戏或者模拟战争策略等. 博弈的基本算法也是模拟人的思维,比如当自己下子时遍历所有可能寻求最有利步 ...

  7. Vi 详细教程

    进入vi的命令 vi filename :打开或新建文件,并将光标置于第一行首 vi +n filename :打开文件,并将光标置于第n行首 vi + filename :打开文件,并将光标置于最后 ...

  8. postfix疯狂外发垃圾邮件

    分析 一.查找main.cf配置文件 localhost# find / -name main.cf /etc/postfix/main.cf 二.打开/etc/postfix/main.cf来看看. ...

  9. vm选项大全

    http://hllvm.group.iteye.com/group/topic/27945 java -XX:后边的总记不住 vm选项大全 http://www.oracle.com/technet ...

  10. 【贪心】时空定位I

    [贪心]时空定位I 题目描述 张 琪曼已经确定了李旭琳在一个长为20千米,宽为2千米的空间,她要在横中心线上放置半径为Ri的定位装置,每个定位装置的效果都会让以它为中心的半径为实 数Ri(0<R ...