AI 助理
备案控制台
开发者社区 开发与运维 文章 正文

java编程思想第四版第十三章字符串 习题

2022-05-08 121
版权
版权声明：
本文内容由阿里云实名注册用户自发贡献，版权归原作者所有，阿里云开发者社区不拥有其著作权，亦不承担相应法律责任。具体规则请查看《 阿里云开发者社区用户服务协议》和 《阿里云开发者社区知识产权保护指引》。如果您发现本社区中有涉嫌抄袭的内容，填写 侵权投诉表单进行举报，一经查实，本社区将立刻删除涉嫌侵权内容。
简介： 运行结果： 运行结果: 输入参数: 输出结果


1.fas


2.第二题


package net.mindview.strings;
import java.util.ArrayList;
import java.util.List;
/**
 * 无限循环
 */
public class InfiniteRecursion {
    public InfiniteRecursion(){
    }
    @Override
    public String toString() {
        return " InfiniteRecursion address" + super.toString() + "\n";
    }
    public static void main(String[] args) {
        List<InfiniteRecursion> list = new ArrayList<InfiniteRecursion>();
        for(int i=0; i<5; i++){
            list.add(new InfiniteRecursion());
        }
        System.out.println(list);
    }
}


3.fa


4.第四题


package net.mindview.strings;
import java.util.Formatter;
public class Receipt {
    private double total = 0;
    private Formatter f = new Formatter(System.out);
    public final int ITEM_WIDTH = 15;
    public final int QTY_WIDTH = 5;
    public final int PRICE_WIDTH = 10;
    public final int PRICISION_FLOAT_WIDHT = 2;
    public final int PRICISION_String_WIDHT = 15;
    public final String TITLE_FORMAT = "%-" + ITEM_WIDTH + "s %" + QTY_WIDTH
            + "s %" + PRICE_WIDTH + "s\n";
    public final String PRICE_FORMAT = "%-" + ITEM_WIDTH + "."
            + PRICISION_String_WIDHT + "s %" + QTY_WIDTH + "d %" + PRICE_WIDTH
            + "." + PRICISION_FLOAT_WIDHT + "f\n"; 
    public final String TOTAL_FORMAT = "%-" + ITEM_WIDTH + "s %" + QTY_WIDTH
            + "s %" + PRICE_WIDTH + "." + PRICISION_FLOAT_WIDHT + "f\n"; 
    //打印标题
    public void printTitle(){
        /*
         * 含义: 格式化字符串串以%开头
         * -: 表示左对齐
         * 15: 15表示宽度
         * s:表示数据的类型是String
         * .2f:表示浮点数保留的小数位数
         * .15s:表示字符串的长度最多15个字符 
         */
        f.format(TITLE_FORMAT, "Item", "Qty", "Price");
        f.format(TITLE_FORMAT, "----", "---", "-----");
    }
    //正文的内容
    public void print(String name, int qty, double price){
        f.format(PRICE_FORMAT, name, qty, price);
        total += price;
    }
    //总价
    public void printTotal(){
        f.format(TOTAL_FORMAT, "Tax", "", total*0.06);
        f.format(TITLE_FORMAT, "", "", "-----");
        f.format(TOTAL_FORMAT, "Total", "", total*1.06);
    }
    public static void main(String[] args) {
        Receipt receipt = new Receipt();
        receipt.printTitle();
        receipt.print("Jack`s Magic Beans", 4, 4.25);
        receipt.print("Princess Peas", 3, 5.1);
        receipt.print("Three Bears Porridge", 1, 14.29);
        receipt.printTotal();
    }
}


运行结果：


Item              Qty      Price
----              ---      -----
Jack`s Magic Be     4       4.25
Princess Peas       3       5.10
Three Bears Por     1      14.29
Tax                         1.42
                           -----
Total                      25.06


5.第五题


package net.mindview.strings;
import java.util.Formatter;
public class Receipt {
    private double total = 0;
    private Formatter f = new Formatter(System.out);
    public final int ITEM_WIDTH = 15;
    public final int QTY_WIDTH = 5;
    public final int PRICE_WIDTH = 10;
    public final int PRICISION_FLOAT_WIDHT = 2;
    public final int PRICISION_String_WIDHT = 15;
    public final String TITLE_FORMAT = "%-" + ITEM_WIDTH + "s %" + QTY_WIDTH
            + "s %" + PRICE_WIDTH + "s\n";
    public final String PRICE_FORMAT = "%-" + ITEM_WIDTH + "."
            + PRICISION_String_WIDHT + "s %" + QTY_WIDTH + "d %" + PRICE_WIDTH
            + "." + PRICISION_FLOAT_WIDHT + "f\n"; 
    public final String TOTAL_FORMAT = "%-" + ITEM_WIDTH + "s %" + QTY_WIDTH
            + "s %" + PRICE_WIDTH + "." + PRICISION_FLOAT_WIDHT + "f\n"; 
    //打印标题
    public void printTitle(){
        /*
         * 含义: 格式化字符串串以%开头
         * -: 表示左对齐
         * 15: 15表示宽度
         * s:表示数据的类型是String
         * .2f:表示浮点数保留的小数位数
         * .15s:表示字符串的长度最多15个字符 
         */
        f.format(TITLE_FORMAT, "Item", "Qty", "Price");
        f.format(TITLE_FORMAT, "----", "---", "-----");
    }
    //正文的内容
    public void print(String name, int qty, double price){
        f.format(PRICE_FORMAT, name, qty, price);
        total += price;
    }
    //总价
    public void printTotal(){
        f.format(TOTAL_FORMAT, "Tax", "", total*0.06);
        f.format(TITLE_FORMAT, "", "", "-----");
        f.format(TOTAL_FORMAT, "Total", "", total*1.06);
    }
    public static void main(String[] args) {
        Receipt receipt = new Receipt();
        receipt.printTitle();
        receipt.print("Jack`s Magic Beans", 4, 4.25);
        receipt.print("Princess Peas", 3, 5.1);
        receipt.print("Three Bears Porridge", 1, 14.29);
        receipt.printTotal();
    }
}


运行结果:


Item              Qty      Price
----              ---      -----
Jack`s Magic Be     4       4.25
Princess Peas       3       5.10
Three Bears Por     1      14.29
Tax                         1.42
                           -----
Total                      25.06


6.afd


7.第七题:


package net.mindview.strings.test7;
public class Test7 {
    public static void main(String[] args) {
        //两种写法都可以
        String regex = "^[A-Z].*\\.$";
        String regex1 = "\\p{Upper}.*\\.$";
        String str = "D.";
        String str1 = "Dfasdfasfasfdasfdasfasfasdf.";
        String str2 = "Dfasdfasfasfdasfdasfasfasdf.E";
        System.out.println(str.matches(regex));
        System.out.println(str1.matches(regex));
        System.out.println(str2.matches(regex));
        System.out.println(str.matches(regex1));
        System.out.println(str1.matches(regex1));
        System.out.println(str2.matches(regex1));
    }
}


运行结果:


true
true
false
true
true
false


8.第八题


package net.mindview.strings;
import java.util.Arrays;
public class Splitting {
    public static String knights = "Then, when you have found the shrubbery, you must cut down the mightiest tree in the forest... with... a herring!";
    public static void split(String regex){
        System.out.println(Arrays.toString(knights.split(regex)));
    }
    public static void main(String[] args) {
        //表示的时按照空格分割字符串
        //运行结果:[Then,, when, you, have, found, the, shrubbery,, you, must, cut, down, the, mightiest, tree, in, the, forest..., with..., a, herring!]
        split(" ");
        //表示按照非单次字符分割字符串--这里的非单次字符是空格和,
        //运行结果:[Then, when, you, have, found, the, shrubbery, you, must, cut, down, the, mightiest, tree, in, the, forest, with, a, herring]
        split("\\W+");
        //这个表示:费单次字符之前带n的地方进行分割字符串 这里的分割符是n空格和n,
        //运行结果:[The, whe, you have found the shrubbery, you must cut dow, the mightiest tree i, the forest... with... a herring!]
        split("n\\W+");
    }
}


package net.mindview.strings.test8;
import net.mindview.strings.Splitting;
public class Test8 {
    public static void main(String[] args) {
        String regex = "the|you";
        Splitting.split(regex);
    }
}


9.第九题


package net.mindview.strings.test9;
import net.mindview.strings.Splitting;
public class Test9 {
    public static void main(String[] args) {
        String regex = "A|E|I|O|U|a|e|i|o|u";
        //通过嵌入式标志表达式  (?i) 也可以启用不区分大小写的匹配。
        String regex1 = "(?i)a|e|i|o|u";
        //[abc] 表示a或b或c
        String regex2 = "(?i)[aeiou]";
        System.out.println(Splitting.knights.replaceAll(regex, "_"));
        System.out.println(Splitting.knights.replaceAll(regex1, "_"));
        System.out.println(Splitting.knights.replaceAll(regex2, "_"));
    }
}


10.第十题


package net.mindview.strings.test10;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test10 {
    public static void main(String[] args) {
        String str = "Java now has regular expressions";
        for(String arg: args){
            Pattern p = Pattern.compile(arg);
            Matcher m = p.matcher(str);
            while(m.find()){
                System.out.println("Match \"" + m.group() +"\" at positions " + m.start() + "-" + (m.end()-1) );
            }
            System.out.println("\n");
        }
    }
}


输入参数:


^Java
\Breg.*
n.w\s+h(a|i)s
s?
s*
s+
s{4}
s{1}.
s{0,3}


输出结果


Match "Java" at positions 0-3
Match "now has" at positions 5-11
Match "" at positions 0--1
Match "" at positions 1-0
Match "" at positions 2-1
Match "" at positions 3-2
Match "" at positions 4-3
Match "" at positions 5-4
Match "" at positions 6-5
Match "" at positions 7-6
Match "" at positions 8-7
Match "" at positions 9-8
Match "" at positions 10-9
Match "s" at positions 11-11
Match "" at positions 12-11
Match "" at positions 13-12
Match "" at positions 14-13
Match "" at positions 15-14
Match "" at positions 16-15
Match "" at positions 17-16
Match "" at positions 18-17
Match "" at positions 19-18
Match "" at positions 20-19
Match "" at positions 21-20
Match "" at positions 22-21
Match "" at positions 23-22
Match "" at positions 24-23
Match "" at positions 25-24
Match "s" at positions 26-26
Match "s" at positions 27-27
Match "" at positions 28-27
Match "" at positions 29-28
Match "" at positions 30-29
Match "s" at positions 31-31
Match "" at positions 32-31
Match "s" at positions 11-11
Match "ss" at positions 26-27
Match "s" at positions 31-31
Match "s " at positions 11-12
Match "ss" at positions 26-27
Match "" at positions 0--1
Match "" at positions 1-0
Match "" at positions 2-1
Match "" at positions 3-2
Match "" at positions 4-3
Match "" at positions 5-4
Match "" at positions 6-5
Match "" at positions 7-6
Match "" at positions 8-7
Match "" at positions 9-8
Match "" at positions 10-9
Match "s" at positions 11-11
Match "" at positions 12-11
Match "" at positions 13-12
Match "" at positions 14-13
Match "" at positions 15-14
Match "" at positions 16-15
Match "" at positions 17-16
Match "" at positions 18-17
Match "" at positions 19-18
Match "" at positions 20-19
Match "" at positions 21-20
Match "" at positions 22-21
Match "" at positions 23-22
Match "" at positions 24-23
Match "" at positions 25-24
Match "ss" at positions 26-27
Match "" at positions 28-27
Match "" at positions 29-28
Match "" at positions 30-29
Match "s" at positions 31-31
Match "" at positions 32-31


11.第十一题


package net.mindview.strings.test11;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test11 {
    public static void main(String[] args) {
        Pattern p = Pattern.compile("(?i)((^[aeiou])|(\\s+[aeiou]))\\w+?[aeiou]\\b");
        Matcher m = p.matcher("Arline ate eight apples and " +
"one orange while Anita hadn't any");
        while(m.find()){
            System.out.println("Match \"" + m.group() +
                    "\" at positions " + m.start() + "-" +
                    (m.end() - 1));
        }
    }
}


运行结果:


Match "Arline" at positions 0-5
Match " ate" at positions 6-9
Match " one" at positions 27-30
Match " orange" at positions 31-37
Match " Anita" at positions 44-49


12.sfa


13.f


14.a


15.fda


16.sdf


17.af


18.a


19.fd


20.af


21.as


22.fd


23.asfd


24.af


25.da


26.f


27.df


28.as


29.fa


30.sf


31.asf


32.a


33.sf


34.af


35.asf


36.


文章标签：
Java
关键词：
Java编程
Java字符串
Java编程思想
Java习题
Java编程思想习题
p2jbaco26vdds
目录
相关文章
yuanzhengme
|
7天前
|
安全 Java 调度
Java编程时多线程操作单核服务器可以不加锁吗？
Java编程时多线程操作单核服务器可以不加锁吗？
yuanzhengme
21 2
1176112968452250
|
10天前
|
Java
死磕-java并发编程技术(二)
死磕-java并发编程技术(二)
1176112968452250
29 1
1176112968452250
|
10天前
|
存储 Java 调度
死磕-java并发编程技术(一）
死磕-java并发编程技术(一）
1176112968452250
26 1
1176112968452250
|
10天前
|
设计模式 缓存 Java
死磕-高效的Java编程(一)
死磕-高效的Java编程(一)
1176112968452250
22 0
刘大猫.
|
5天前
|
Java 数据库
java小工具util系列1：日期和字符串转换工具
java小工具util系列1：日期和字符串转换工具
刘大猫.
15 3
君子世无双~
|
11天前
|
存储 Java
Java编程中的对象和类
【8月更文挑战第55天】在Java的世界中，“对象”与“类”是构建一切的基础。就像乐高积木一样，类定义了形状和结构，而对象则是根据这些设计拼装出来的具体作品。本篇文章将通过一个简单的例子，展示如何从零开始创建一个类，并利用它来制作我们的第一个Java对象。准备好让你的编程之旅起飞了吗？让我们一起来探索这个神奇的过程！
君子世无双~
25 10
拉丁解牛说技术
|
12天前
|
Java API 容器
JAVA并发编程系列(10)Condition条件队列-并发协作者
本文通过一线大厂面试真题，模拟消费者-生产者的场景，通过简洁的代码演示，帮助读者快速理解并复用。文章还详细解释了Condition与Object.wait()、notify()的区别，并探讨了Condition的核心原理及其实现机制。
拉丁解牛说技术
32 11
拉丁解牛说技术
|
12天前
|
Java
JAVA并发编程系列(9)CyclicBarrier循环屏障原理分析
本文介绍了拼多多面试中的模拟拼团问题，通过使用 `CyclicBarrier` 实现了多人拼团成功后提交订单并支付的功能。与之前的 `CountDownLatch` 方法不同，`CyclicBarrier` 能够确保所有线程到达屏障点后继续执行，并且屏障可重复使用。文章详细解析了 `CyclicBarrier` 的核心原理及使用方法，并通过代码示例展示了其工作流程。最后，文章还提供了 `CyclicBarrier` 的源码分析，帮助读者深入理解其实现机制。
拉丁解牛说技术
38 11
shuj
|
12天前
|
设计模式 安全 Java
Java 编程中的设计模式：单例模式的深度解析
【9月更文挑战第22天】在Java的世界里，单例模式就像是一位老练的舞者，轻盈地穿梭在对象创建的舞台上。它确保了一个类仅有一个实例，并提供全局访问点。这不仅仅是代码优雅的体现，更是资源管理的高手。我们将一起探索单例模式的奥秘，从基础实现到高级应用，再到它与现代Java版本的舞蹈，让我们揭开单例模式的面纱，一探究竟。
shuj
23 11
刘大猫.
|
5天前
|
SQL Java 索引
java小工具util系列2：字符串工具
java小工具util系列2：字符串工具
刘大猫.
8 2

热门文章

最新文章

  • 1
    java.io.IOException: Stream closed解决办法
  • 2
    Java获取URL中的顶级域名domain的工具类
  • 3
    阿里Java代码规约插件即将全球首发，邀您来发布仪式现场
  • 4
    【Java核心技术卷】深入理解Java的内部类
  • 5
    《Java 7程序设计入门经典》一1.18　习题
  • 6
    Java创建透明与任意形状窗口
  • 7
    一道面试题考验了你对java的理解程度
  • 8
    Java内存模型
  • 9
    单链表_JAVA描述《数据结构与算法分析》
  • 10
    JAVA之旅（九）——Object类，equals，toString，getClass，内部类访问规则，静态内部类,内部类原则，匿名内部类
  • 1
    构建高性能微服务架构：Java后端的实践之路
    90
  • 2
    Java 8新特性之Lambda表达式
    38
  • 3
    windows 下 win11 JDK17安装与环境变量的配置(配置简单详细,包含IJ中java文件如何使用命令运行)
    1637
  • 4
    Java基础语法之基本语法格式
    35
  • 5
    构建高效Android应用：探究Kotlin与Java的性能对比
    75
  • 6
    Java集合的分类有哪些？
    83
  • 7
    java基于云部署的SaaS医院云HIS系统源码 心理CT、B超 lis、电子病历
    89
  • 8
    Java 并发编程的探索与实践
    54
  • 9
    Maven 构建 Java 项目
    61
  • 10
    18：面向切面编程-Java Spring
    123

    • 相关课程

    更多
  • Java面试疑难点解析 - 面试技巧及语言基础
  • Java面试疑难点解析 - 系统架构及项目设计
  • Java编程入门
  • Java面向对象编程
  • Java高级编程

    • 相关电子书

    更多
  • Spring Cloud Alibaba - 重新定义 Java Cloud-Native
  • The Reactive Cloud Native Arch
  • JAVA开发手册1.5.0

    • 相关实验场景

    更多
  • 基于ECS搭建Java Web开发环境
  • 使用Aliyun Java Intializr进行项目开发
  • 搭建Java Web开发环境
  • 部署基于Dragonwell的Java运行环境
  • 阿里云平台上进行Java程序的编译与运行
    • 下一篇
    无影云桌面