1

该程序允许用户输入一个短语并将其转换为 ROT13，其中输入的每个英文字母在其后 13 位变为字母（A 变为 N）。我当前的代码在输入 1 个字符时有效，但是我需要它在代码中运行字符的次数。我试图在开始时放入一个while循环，但它似乎不起作用。为什么是这样？

import java.io.*;

public class J4_1_EncryptionErasetestCNewTry
{

    public static void main (String [] args) throws IOException
    {
        BufferedReader myInput = new BufferedReader (new InputStreamReader (System.in));// Buffered Reader reads the number inputed 

        String key [] = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
        String keyA [] = {"N","O","P","Q","R","S","T","U","V","W","X","Y","Z","A","B","C","D","E","F","G","H","I","J","K","L","M"};

        System.out.println("Enter a phrase: ");
        String phrase = myInput.readLine();

        int length = phrase.length();
        int y = 0, i = 0, num = 0;

        while (y <= length) {
            String letter = Character.toString(phrase.charAt(y));
            y++;
            while(!(letter.equals(key[i]))){
                i++;
            }
            num = i;
            System.out.println(keyA[num]);
            y++;
        }
    }
}
4

5 回答 5

1

请参阅代码注释。

public static void main(String[] args) {

        BufferedReader myInput = new BufferedReader (new InputStreamReader (System.in));// Buffered Reader reads the number inputed 

        String key [] = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
        String keyA [] = {"N","O","P","Q","R","S","T","U","V","W","X","Y","Z","A","B","C","D","E","F","G","H","I","J","K","L","M"};

        System.out.println("Enter a phrase: ");
        String phrase = "";

        try {
            phrase = myInput.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }

        int length = phrase.length();
        int y = 0, i = 0, num = 0;

        while (y < length) { // This should be y < length. Otherwise, it would throw a StringIndexOutOfBoundsException.
            i=0; // Re-initialize
            String letter = Character.toString(phrase.charAt(y));
//            y++; // Unecessary incremental
            while(!(letter.equalsIgnoreCase(key[i]))){
                i++;
            }
            num = i;
            System.out.print(keyA[num]);
            y++;
        }

    }
于 2013-09-13T01:58:25.097 回答
1

虽然这不能回答你的问题，但它回答了你的意图：

public static String rot13(String s) {
    String r = "";
    for (byte b : s.getBytes())
        r += (char)((b + 13 - 'A') % 26 + 'A');
    return r;
}

您的代码对于它正在做的事情来说太复杂了。真的，所有的工作都可以在一条线上完成。使用字节算术而不是数组查找等。简单/更少的代码始终是最好的方法。

请不要对效率低下等发表评论。这是一个有效的基本实现（经过测试）。读者可以自由地改进它作为练习。

于 2013-09-13T02:24:29.167 回答
0

我以不同的方式实现了它，但它可以按您的预期工作，仅适用于您的示例中的大写：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.List;

public class WhileLoopIssue {

    public static void main( String[] args ) throws IOException {
        BufferedReader myInput = new BufferedReader( new InputStreamReader(
                System.in ) );// Buffered Reader reads the
                              // number inputed

        final List<String> letterList = Arrays.asList( "A", "B", "C", "D", "E",
                "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q",
                "R", "S", "T", "U", "V", "W", "X", "Y", "Z" );

        System.out.println( "Enter a phrase: " );
        String phrase = myInput.readLine();

        final String[] letters = phrase.split( "" );     // Split input phrase
        final StringBuffer buffer = new StringBuffer();  // Variable to save letters. Could be a String as well.
        for ( int i = 0; i < letters.length; i++ ) {
            final int letterIndex = letterList.indexOf( letters[i] );  // Get the numeric value of the letter
            if ( letterIndex < 0 )  // Skip iteration if not found. Maybe a lowercase, or an empty String
                continue;

            final int nextLetterIndex = 13 + letterIndex;   // Actual value of the letter + 13
            if ( nextLetterIndex > letterList.size() ) {
                buffer.append( nextLetterIndex % letterList.size() );  // Actual value greater than the total number of letters in the alphabet, so we get the modulus for the letter
            } else {
                buffer.append( letterList.get( nextLetterIndex ) );  // Letter in the range, get it
            }
        }
        System.out.println( buffer.toString() );
    }
}
于 2013-09-13T01:46:22.167 回答
0

您的代码很可能会在您的内部 while 循环中中断，因为您没有重置i. 如果不这样做，您将遇到 StringIndexOutOfBounds。我建议i在您的外部 while 循环中进行初始化，或者更好地int i = 0;在外部 while 循环中移动。

于 2013-09-13T01:48:32.930 回答
0

您需要在每次迭代中重置 i 。可能会在数组“key”的末尾找到第一个字母。您的代码将在那里找到下一个输入字符，我想这不是您想要的，并且不会找到该字符并会抛出 SIOBException。我已经更改了 while 循环，也删除了变量 y 中的两次增量。看一看

    while (y < length) {
        i = 0; //Every Time you want to search from start of the array 
                //so just reset the i.
        String letter = Character.toString(phrase.charAt(y));
        while(!(letter.equals(key[i]))){
            i++;
        }
        num = i;
        System.out.println(keyA[num]);
        y++;
    }

我假设您输入的任何内容都是仅包含大写字母的短语，否则您将遇到 SIOBException，因为您将无法在数组中找到该字母。

顺便说一句，您应该使用其他一些对搜索有效的数据结构，而不是那些数组，例如 hashmap。您在数组中的线性搜索未优化。

于 2013-09-13T01:50:51.637 回答