Question

实施方法
public void search (String searchString) { }
迭代注释ArrayList直到它找到包含searchString的注释。然后它应该打印找到的项目或消息“未找到字符串”。

到目前为止，我有：

import java.util.ArrayList;
import java.util.Iterator;

/**
 * A class to maintain an arbitrarily long list of notes.
 * Notes are numbered for external reference by a human user.
 * In this version, note numbers start at 0.
 * 
 * @author David J. Barnes and Michael Kolling.
 * @version 2008.03.30
 */
public class Notebook
{
    // Storage for an arbitrary number of notes.
    private ArrayList<String> notes;

    /**
     * Perform any initialization that is required for the
     * notebook.
     */
    public Notebook()
    {
        notes = new ArrayList<String>();
    }

    /**
     * Store a new note into the notebook.
     * @param note The note to be stored.
     */
    public void storeNote(String note)
    {
        notes.add(note);
    }

    /**
     * @return The number of notes currently in the notebook.
     */
    public int numberOfNotes()
    {
        return notes.size();
    }

    /**
     * Show a note.
     * @param noteNumber The number of the note to be shown.
     */
    public void showNote(int noteNumber)
    {
        if(noteNumber < 0) {
            // This is not a valid note number, so do nothing.
            System.out.println("invalid index given");
        }
        else if(noteNumber < numberOfNotes()) {
            // This is a valid note number, so we can print it.
            System.out.println(notes.get(noteNumber));
        }
        else {
             System.out.println("there are fewer items in the notebook");
            // This is not a valid note number, so do nothing.
        }
    }

    public void removeNote(int noteNumber)
    {
        if(noteNumber < 0) {
            // This is not a valid note number, so do nothing.
             System.out.println("invalid index given");
        }
        else if(noteNumber < numberOfNotes()) {
            // This is a valid note number.
            notes.remove(noteNumber);
        }
        else {
            System.out.println("there are fewer items in the notebook");
            // This is not a valid note number, so do nothing.
        }
    }

    /* Edit note.
     * I tried to improve the formatting of the code below, but I'm completely
     * unable to figure out how on earth anything of that should make sense
     * and therefore the indentation is completely without any meaning.
     */
    public void search (String searchString)
    { 
        for each notes in ArrayList {    
            if notes = searchString;        
                System.out.println("String found"); + searchString        
                return    end 
            }
        if}
        System.out.println("String not found");
    }
}

但它不起作用，我无法解决这个问题。

Answer 1

几个问题：

您的search方法实际上在课堂外。
您的搜索方法正文根本没有意义。

如果你依靠写莎士比亚的猴子，你会等待一段时间。

Answer 2

要迭代数组列表，您可以使用'for-each' loop：

for (String note: notes) {
    // Do something with note
}

这是非常基本的语法。你以前见过吗？如果没有，您应该在尝试这项功课之前先阅读一本非常基本的Java教程。

Answer 3

从根本上说，您需要查看ArrayList中的每个项目并测试它是否与搜索条件匹配。在伪代码中

for each note in notes
{
    if note equals searchString then 
        print "Found " + searchString
        return
    end if
}

print "not found"

鉴于基本概要，想在Java中编写它的第二次尝试？

Answer 4

基本上你想循环遍历元素，并为每个元素检查它是否等于你要搜索的元素。您可以使用for循环或foreach循环来进行实际迭代。

Answer 5

你应该检查整个音符是否与你的搜索字符串匹配，或者音符是否包含你的搜索字符串？

即。给出“foobar”，“baz”，“垃圾邮件”的注释，如果搜索“foo”返回“foobar”或不匹配任何内容？

所以在伪代码中：

for each note in notes
{
     if searchstring in note
     {
           print "Found :"+note
     }
}

Answer 6

查看此网站http://pleac.sourceforge.net/pleac_java/arrays.html它可能有用

搜索ArrayList

6 个答案: