[LeetCode OJ] Max Points on a Line
Submission Details
|
27 / 27 test cases passed.
|
Status:
Accepted |
|
Runtime: 472 ms
|
Submitted: 0 minutes ago
|
Submitted Code
/**
* Definition for a point.
* class Point {
* int x;
* int y;
* Point() { x = 0; y = 0; }
* Point(int a, int b) { x = a; y = b; }
* }
*/
import java.util.HashMap;
import java.util.Map; public class Solution {
public int maxPoints(Point[] points) {
if (points == null || points.length == 0)
return 0;
int N = points.length;
if (N == 1)
return 1;
if (N == 2)
return 2; Map<String, Integer> dict = new HashMap<String, Integer>();
int maxOverall = 0;
for (int i = 0; i < N - 1; i++) {
int same = 1;
int max = 0;
for (int j = i + 1; j < N; j++) { int p1x = points[i].x;
int p1y = points[i].y;
int p2x = points[j].x;
int p2y = points[j].y; if (p1x == p2x && p1y == p2y) {
same++;
continue;
} int A = p2y - p1y;
int B = p1x - p2x; int GCD = gcd(A, B);
if (GCD != 0 && GCD != 1) {
A /= GCD;
B /= GCD;
}
if (A < 0) {
A = -A;
B = -B;
} String key = A + "," + B;
// System.out.println("round:" + i + " " + key);
int value = 1;
if (dict.containsKey(key)) {
value += dict.get(key);
}
dict.put(key, value);
max = max < value ? value : max;
} max += same;
maxOverall = maxOverall < max ? max : maxOverall;
dict.clear();
} return maxOverall;
} public int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
}
[LeetCode OJ] Max Points on a Line的更多相关文章
- [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 ...
- 【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 ...
- [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. ...
- [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. ...
- 【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. ...
- 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. ...
- 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. ...
- 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 ...
- 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
- SecWeek
“叮铃铃,叮铃铃”清晨闹钟如期而至,每天的这个时候都会是一天中最头痛的时刻,每日坚持早起,渐渐已经开始习惯,扶着沉重的额头,侧身翻起,会觉得世界都在天旋地转. 一个人,悄悄的开门,悄悄的刷牙洗脸,然后 ...
- CoreData数据库
一 CoreData 了解 1 CoreData 数据持久化框架是 Cocoa API 的一部分,首先在iOSS5 版本的系统中出现: 它允许按照 实体-属性-值 模式组织数据: ...
- IOS 二维码的实现
1.首先导入Coreimage框架. //创建滤镜对象 CIFilter *filter = [CIFilter filterWithName:@"CIQRCodeGenerator&quo ...
- commonJS
摘自阮一峰博客:http://javascript.ruanyifeng.com/nodejs/module.html 目录 概述 module对象 module.exports属性 exports变 ...
- Silverlight RadChart :创建十字定位&圈选
//图像加载 void Chart_Loaded(object sender, RoutedEventArgs e) { var plotAreaPanel = this.radChart.Defau ...
- C/C++文字常量与常变量的概念与区别 分类: C/C++ 2015-06-10 22:56 111人阅读 评论(0) 收藏
以下代码使用平台是Windows 64bits+VS2012. 在C/C++编程时,经常遇到以下几个概念:常量.文字常量.符号常量.字面常量.常变量.字符串常量和字符常量,网上博客资料也是千篇千律,不 ...
- IOS上解决内存越界访问问题
IOS经常会混合使用C代码,而在C中,对内存的读写是很频繁的操作. 其中,内存越界读写 unsigned char* p =(unsigned char*)malloc(10); unsigned c ...
- jquey easyui 常用方法
jquey easyui 常用方法 2015-05-31 13:02 4473人阅读 评论(0) 收藏 举报 版本:1.4.2 一.easyui -textbox: 1.去空格: $('#tt1'). ...
- Linux LVM硬盘管理之一:概念介绍
一.LVM概念介绍: LVM是 Logical Volume Manager(逻辑卷管理)的简写,它由Heinz Mauelshagen在Linux 2.4内核上实现.LVM将一个或多个硬盘的分区在逻 ...
- 崽崽帮www.zaizaibang.com精选2
崔其亮-儿科专科 智慧熊北京丰台幼儿园——秋游活动 周边景点——哈尔滨旅游攻略 月是故乡明丨那些买给自己吃的常德手工月饼老店 [周末乐游]南京最美10所大学,这个周末一起逛一逛吧 彭韶-儿科二门诊 [ ...