Java练习20题

第一关：Welcome to Java

You must print two lines of output:

1. Print Hello, World. on the first line.
2. Print Hello, Java. on the second line.

解决方案：

public class Solution {
    public static void main(String[] args) {
        System.out.println("Hello, World.");
        System.out.println("Hello, Java.");
    }
}

第二关：Java Stdin and Stdout I

用户输入三个整数，打印出来。

Sample Input

42
100
125

Sample Output

42
100
125

解决方案：

public class Solution {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int a = scan.nextInt();
        int b = scan.nextInt();
        int c = scan.nextInt();
        scanner.close();
        System.out.println(a);
        System.out.println(b);
        System.out.println(c);
    }
}

第三关：If-Else

任务：

输入一个整数n，如果是奇数，输出Weird；如果是[2,5]的偶数，输出Not Weird，[6,20]的偶数，输出Weird，大于20的偶数，输出Not Weird

解决方案：

public class Solution {
    private static final Scanner scanner = new Scanner(System.in);
    public static void main(String[] args) {
        int N = scanner.nextInt();
        scanner.skip("(
|[

u2028u2029u0085])?");
        if(N%2==1) {
            System.out.println("Weird");
        } else {
            if((N>=2 && N<=5) || (N>20)) {
                System.out.println("Not Weird");
            } else {
                System.out.println("Weird");
            }
        }
        scanner.close();
    }
}

里面有句：

scanner.skip("(
|[

u2028u2029u0085])?");

以下解释来自stackoverflow：

• ? 匹配0或1个前面的字符
• | 选择
• [] 匹配其中的单个字符
• u2028 行分隔符
• u2029 段分隔符
• u0085 下一行

总结一下就是跳过 (Windows 行结束)或 u2028u2029u0085的其中一个。

举例：对于代码

int a  =scanner.nextInt();
String s = scanner.nextLine();

用户输入：

4
This is next line

s的结果是空串。

用户输入：

4 This is next line

s的结果是This is next line。所以通过skip跳过这些换行之类的操作，nextline的字符串就正确了。

第四关：Java Stdin and Stdout II

Sample Input

42
3.1415
Welcome to HackerRank's Java tutorials!

Sample Output

String: Welcome to HackerRank's Java tutorials!
Double: 3.1415
Int: 42

基本就是上面的如何跳过nextLine之前的换行。

public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    int i = scan.nextInt();
    double d = scan.nextDouble();
    scan.skip("(
|[

u2028u2029u0085])?");
    String s = scan.nextLine();

    System.out.println("String: " + s);
    System.out.println("Double: " + d);
    System.out.println("Int: " + i);
}

第五关：Java Output Formatting

输出排版，数字是第15个字符，不足三位补0

Sample Input

java 100
cpp 65
python 50

Sample Output

================================
java           100 
cpp            065 
python         050 
================================

解决方案：

import java.util.Scanner;
import java.util.Formatter;

public class Solution {
    static Formatter formatter = new Formatter(System.out);
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        System.out.println("================================");
        for(int i=0;i<3;i++)
        {
            String s1=sc.next();
            int x=sc.nextInt();
            formatter.format("%-14s %03d
",s1,x);
        }
        System.out.println("================================");
        sc.close();

    }
}

第六关：Java Loops I

输入一个数，打印它和[1,10]的乘积表达式。

Sample Input

2

Sample Output

2 x 1 = 2
2 x 2 = 4
2 x 3 = 6
2 x 4 = 8
2 x 5 = 10
2 x 6 = 12
2 x 7 = 14
2 x 8 = 16
2 x 9 = 18
2 x 10 = 20

解决方案：

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        //  readLine()方法会返回用户在按下Enter键之前的所有字符输入,不包括最后按下的Enter返回字符
        int N = Integer.parseInt(bufferedReader.readLine().trim());
        for(int i=1;i<=10;i++) {
            System.out.printf("%d x %d = %d
",N,i,N*i);
        }
        bufferedReader.close();
    }
}

第七关：Java Loops II

第一行输入计算的次数，每次计算输入a, b, n，输出：

Sample Input

2
0 2 10
5 3 5

Sample Output

2 6 14 30 62 126 254 510 1022 2046
8 14 26 50 98

解决方案：

public static void main(String []argh){
    Scanner in = new Scanner(System.in);
    int t=in.nextInt();
    for(int i=0;i<t;i++){ int="" a="in.nextInt();" b="in.nextInt();" n="in.nextInt();" result="a;" pow="1;" for(int="" j="0;j<n-1;j++)" {="" +="pow*b;" system.out.printf("%d="" ",result);="" pow*="2;" }="" system.out.printf("%d
",result+pow*b);="" in.close();="" ```="" ##="" 第八关：java="" int数据类型="" java整数类型：="" -="" bytes="" 8位有符号整型="" short="" 16位有符号整型="" 32位有符号整型="" long="" 64位有符号整型="" 输入一个数，判断能是什么类型。="" **sample="" input**="" ```shell="" 5-1501500001500000000213333333333333333333333333333333333-100000000000000="" output**="" -150="" can="" be="" fitted="" in:*="" short*="" int*="" long150000="" long1500000000="" long213333333333333333333333333333333333="" can't="" anywhere.-100000000000000="" **解决方案：**="" ```java="" public="" static="" void="" main(string="" []argh){="" scanner="" sc="new" scanner(system.in);="" t="sc.nextInt();" i="0;i<t;i++)" try="" x="sc.nextLong();" system.out.println(x+"="" in:");="" if(x="">=Byte.MIN_VALUE && x<=Byte.MAX_VALUE)System.out.println("* byte");            if(x>=Short.MIN_VALUE && x<=Short.MAX_VALUE)System.out.println("* short");            if(x>=Integer.MIN_VALUE && x<=Integer.MAX_VALUE)System.out.println("* int");            if(x>=Long.MIN_VALUE && x<=Long.MAX_VALUE)System.out.println("* long");        }        catch(Exception e)        {            System.out.println(sc.next()+" can't be fitted anywhere.");        }    }}

用try catch判断是否能当做整数读入，能再继续判断。

第九关：Java End-of-file

统计读入EOF之前的行数。

Sample Input

Hello worldI am a fileRead me until end-of-file.

Sample Output

1 Hello world2 I am a file3 Read me until end-of-file.

解决方案：

public static void main(String[] args) throws IOException{    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));    String s;    int n = 1;    while((s=br.readLine())!=null) {        System.out.println(n+" "+s);        n+=1;    }    br.close();}

第十关：Java Static Initializer Block

输入长和宽，如果都为正，输出乘积，否则打印一个字符串。

public class Solution {    static Scanner scanner = new Scanner(System.in);    static int B = scanner.nextInt();    static int H = scanner.nextInt();    static boolean flag = false;    static {        if(B>0&&H>0) {            flag = true;        } else {            System.out.println("java.lang.Exception: Breadth and height must be positive");        }    }	public static void main(String[] args){		if(flag){			int area=B*H;			System.out.print(area);		}    }}

类方法的执行顺序：

public class Demo extends Base{    static {        System.out.println("test static");    }    public Demo(){        System.out.println("test constructor");    }    public static void main(String[] args) {        System.out.println("Demo main func start");        new Demo();        System.out.println("Demo main func end");    }}class Base{    static{        System.out.println("base static");    }    public Base(){        System.out.println("base constructor");    }}

输出

base statictest staticDemo main func startbase constructortest constructorDemo main func end

程序执行，先找到main方法，执行main前发现继承了Base类，就先执行Base的静态方法，然后是Demo的静态方法。之后开始执行main函数，创建了一个新对象Demo，先执行父类Base的构造函数，然后是Demo的构造函数，之后继续执行main方法。

第十一关：Java Int to String

一行：

String s = String.valueOf(n);

第十二关：Java Date and Time

输入年月日，输出星期几。

Sample Input

08 05 2015

Sample Output

WEDNESDAY

解决方案：

public static String findDay(int month, int day, int year) {
    LocalDate dt = LocalDate.of(year, month, day);
    return dt.getDayOfWeek().toString();
}

LocalDate的使用：

源自：java8 — 新日期时间API篇

Java8出的新的时间日期API都是线程安全的，并且性能更好，代码更简洁！

新时间日期API常用重要对象介绍：

• ZoneId: 时区ID，用来确定Instant和LocalDateTime互相转换的规则
• Instant: 用来表示时间线上的一个点（瞬时）
• LocalDate: 表示没有时区的日期, LocalDate是不可变并且线程安全的
• LocalTime: 表示没有时区的时间, LocalTime是不可变并且线程安全的
• LocalDateTime: 表示没有时区的日期时间, LocalDateTime是不可变并且线程安全的
• Clock: 用于访问当前时刻、日期、时间，用到时区
• Duration: 用秒和纳秒表示时间的数量（长短），用于计算两个日期的“时间”间隔
• Period: 用于计算两个日期间隔

应用示例：

LocalDate localDate = LocalDate.now(); // 年月日
System.out.println(localDate); // 2021-10-11

LocalDateTime localDateTime = LocalDateTime.now(); // 具体时间
System.out.println(localDateTime); //  2021-10-11T21:20:28.455

LocalDate localDate1 = LocalDate.of(2021,10,11);
System.out.println(localDate1.getDayOfWeek()); // MONDAY
LocalDate localDate2 = localDate1.plusDays(5); // 加上5天
System.out.println(localDate2.getDayOfWeek()); // SATURDAY
System.out.println(localDate1.isLeapYear()); // 是否是闰年 false

// 格式化
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy年MM月dd日 HH:mm:ss");
System.out.println(formatter.format(localDateTime)); // 时间格式化 2021年10月11日 21:31:29

// Period 时间间隔
Period period = Period.between(localDate1,localDate2); //时间间隔应为5天
System.out.println(period.getDays()); // 5

第十三关：Java Currency Formatter

Sample Input

12324.134

Sample Output

US: $12,324.13
India: Rs.12,324.13
China: ￥12,324.13
France: 12 324,13 €

NumberFormat有一个专门处理货币的类。

public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    double payment = scanner.nextDouble();
    scanner.close();

    NumberFormat numberFormat = NumberFormat.getCurrencyInstance(Locale.US);
    String us = numberFormat.format(payment);

    numberFormat = NumberFormat.getCurrencyInstance(new Locale("en","IN"));
    String india = numberFormat.format(payment);

    numberFormat = NumberFormat.getCurrencyInstance(Locale.CHINA);
    String china = numberFormat.format(payment);

    numberFormat = NumberFormat.getCurrencyInstance(Locale.FRANCE);
    String france = numberFormat.format(payment);


    System.out.println("US: " + us);
    System.out.println("India: " + india);
    System.out.println("China: " + china);
    System.out.println("France: " + france);
}

印度怎么和其它的不一样？

第十四关：Java Sort

ID NAME SCORE，先按SCORE逆序，再按名字正序。

Sample Input

533 Rumpa 3.6885 Ashis 3.8556 Samiha 3.7519 Samara 3.7522 Fahim 3.76

Sample Output

Ashis
Fahim
Samara
Samiha
Rumpa

解决方案：

import java.util.*;

class Student{
	private int id;
	private String fname;
	private double cgpa;
	public Student(int id, String fname, double cgpa) {
		super();
		this.id = id;
		this.fname = fname;
		this.cgpa = cgpa;
	}
	public int getId() {
		return id;
	}
	public String getFname() {
		return fname;
	}
	public double getCgpa() {
		return cgpa;
	}
}

public class Solution
{
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		int testCases = Integer.parseInt(in.nextLine());
		
		List<student> studentList = new ArrayList<student>();
		while(testCases>0){
			int id = in.nextInt();
			String fname = in.next();
			double cgpa = in.nextDouble();
			
			Student st = new Student(id, fname, cgpa);
			studentList.add(st);
			
			testCases--;
		}
        
        // ========== 排序代码
        
      	for(Student st: studentList){
			System.out.println(st.getFname());
		}
	}
}

写法一：

Collections.sort(studentList, new Comparator<student>() {
    @Override
    public int compare(Student o1, Student o2) {
        if (o1.getCgpa()==o2.getCgpa()) {
            return o1.getFname().compareTo(o2.getFname());
        }
        return Double.compare(o2.getCgpa(),o1.getCgpa());
    }
});

写法二：

Collections.sort(studentList, (o1, o2) -> {
    if (o1.getCgpa()==o2.getCgpa()) {
        return o1.getFname().compareTo(o2.getFname());
    }
    return Double.compare(o2.getCgpa(),o1.getCgpa());
});

写法三：

studentList.sort((o1, o2) -> {
    if (o1.getCgpa() == o2.getCgpa()) {
        return o1.getFname().compareTo(o2.getFname());
    }
    return Double.compare(o2.getCgpa(), o1.getCgpa());
});

第十五关：Java Dequeue

滑窗unique number的最大数，双端队列。

Sample Input

6 3 // 6是6个数，3是窗口大小为3
5 3 5 2 3 2

Sample Output

3

即5 3 2, 3 5 3, 5 2 3，2 3 2其中uniq number最多为3。

解决方案：

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    Deque<integer> deque = new ArrayDeque<>();
    Map<integer,integer> map = new HashMap<>();
    int nn = in.nextInt();
    int m = in.nextInt();

    for (int i = 0; i < m; i++) {
        int num = in.nextInt();
        deque.addLast(num);
        map.put(num,map.getOrDefault(num,0)+1);
    }
    int n = map.size(); // cur unique num length
    int maxNum = n; // max uniq num len
    int first;
    for (int i = m; i < nn; i++) {
        int num = in.nextInt();
        first = deque.pollFirst();
        map.put(first,map.get(first)-1);
        if(map.get(first)==0) {
            n -= 1;
        }
        deque.addLast(num);
        map.put(num,map.getOrDefault(num,0)+1);
        if(map.get(num)==1) {
            n += 1;
        }
        maxNum = n>maxNum?n:maxNum;
    }
    System.out.println(maxNum);
}

第十六关：返回的类型

看看就好。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

//Complete the classes below
class Flower {
    public String whatsYourName() {
        return "I have many names and types.";
    }
}

class Jasmine extends Flower {
    @Override
    public String whatsYourName() {
        return "Jasmine";
    }
}

class Lily extends Flower {
    @Override
    public String whatsYourName() {
        return "Lily";
    }
}

class Region {
    public Flower yourNationalFlower() {
        return new Flower();
    } 
}

class WestBengal extends Region{
    @Override
    public Flower yourNationalFlower() {
        return new Jasmine();
    } 
}

class AndhraPradesh  extends Region{
    @Override
    public Flower yourNationalFlower() {
        return new Lily();
    } 
}


public class Solution {
  public static void main(String[] args) throws IOException {
      BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
      String s = reader.readLine().trim();
      Region region = null;
      switch (s) {
        case "WestBengal":
          region = new WestBengal();
          break;
        case "AndhraPradesh":
          region = new AndhraPradesh();
          break;
      }
      Flower flower = region.yourNationalFlower();
      System.out.println(flower.whatsYourName());
    }
}

第十七关：Java Method Overriding

调用父类被重写的方法。输出：

Hello I am a motorcycle, I am a cycle with an engine.
My ancestor is a cycle who is a vehicle with pedals.

解决方案：

import java.util.*;
import java.io.*;


class BiCycle{
	String define_me(){
		return "a vehicle with pedals.";
	}
}

class MotorCycle extends BiCycle{
	String define_me(){
		return "a cycle with an engine.";
	}
	
	MotorCycle(){
		System.out.println("Hello I am a motorcycle, I am "+ define_me());
		String temp=super.define_me(); //Fix this line
		System.out.println("My ancestor is a cycle who is "+ temp );
	}
}
class Solution{
	public static void main(String []args){
		MotorCycle M=new MotorCycle();
	}
}

第十八关：Java Instanceof keyword

Object element=mylist.get(i);
if(element instanceof Student)
    ...

第十九关：Java Annotations

元注解：

@Target(ElementType.METHOD) // 用到方法上
@Retention(RetentionPolicy.RUNTIME) // 运行时生效
@interface FamilyBudget {
   String userRole() default "GUEST";
}

详见Java反射三四例

大致题意就是判断花费是否超预算（注解形式）：

import java.lang.annotation.*;
import java.lang.reflect.*;
import java.util.*;

@Target(ElementType.METHOD)
@Retention(RetentionPolicy.RUNTIME)
@interface FamilyBudget {
	String userRole() default "GUEST";
	//~~Complete the interface~~
    int budgetLimit() default 0;
}

class FamilyMember {
    //~~Complete this line~~
	@FamilyBudget(userRole = "SENIOR",budgetLimit=100)
	public void seniorMember(int budget, int moneySpend) {
		System.out.println("Senior Member");
		System.out.println("Spend: " + moneySpend);
		System.out.println("Budget Left: " + (budget - moneySpend));
	}

	//~~Complete this line~~
    @FamilyBudget(userRole = "JUNIOR",budgetLimit=50)
	public void juniorUser(int budget, int moneySpend) {
		System.out.println("Junior Member");
		System.out.println("Spend: " + moneySpend);
		System.out.println("Budget Left: " + (budget - moneySpend));
	}
}

public class Solution {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int testCases = Integer.parseInt(in.nextLine());
		while (testCases > 0) {
			String role = in.next();
			int spend = in.nextInt();
			try {
				Class annotatedClass = FamilyMember.class;
				Method[] methods = annotatedClass.getMethods();
				for (Method method : methods) {
					if (method.isAnnotationPresent(FamilyBudget.class)) {
						FamilyBudget family = method
								.getAnnotation(FamilyBudget.class);
						String userRole = family.userRole();
						//int budgetLimit = ~~Complete this line~~;
                        int budgetLimit = family.budgetLimit();
						if (userRole.equals(role)) {
							//if(~~Complete this line~~){
                            if(budgetLimit>=spend) {
								method.invoke(FamilyMember.class.newInstance(),
										budgetLimit, spend);
							}else{
								System.out.println("Budget Limit Over");
							}
						}
					}
				}
			} catch (Exception e) {
				e.printStackTrace();
			}
			testCases--;
		}
	}
}

第二十关：Java BigDecimal

大数逆序排序

Sample Input

9  //个数
-100
50
0
56.6
90
0.12
.12
02.34
000.000

Sample Output

90
56.6
50
02.34
0.12
.12
0
000.000
-100

解决方案：

public static void main(String[] args) {
    //Input
    Scanner sc= new Scanner(System.in);
    int n=sc.nextInt();
    String []s=new String[n];
    for(int i=0;i<n;i++){ s[i]="sc.next();" }="" sc.close();="" arrays.sort(s,="" collections.reverseorder((a1,="" a2)="" -=""> {
        BigDecimal a = new BigDecimal(a1);
        BigDecimal b = new BigDecimal(a2);
        return a.compareTo(b);
    }));

    //Output
    for(int i=0;i<n;i++)
    {
        System.out.println(s[i]);
    }
}

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