我正在尝试创建一个将英语转换为自定义二进制文件的翻译器，反之亦然。它将英语翻译成二进制，但不是二进制翻译成英语。请有人帮助我，告诉我我做错了什么？
我想输入类似“01000 00101 01100 01100 01111”的内容并获得“HELLO”作为输出。现在程序只允许我一次输入一个字符。例如，我输入“01000”，我得到了h。但是当我输入类似“01000 00101”的东西时，我得到同样的回报。我想知道如何解决这个问题。
以下是代码：
import java.util.Scanner;
public class Translator
{
public static void main ( String [] args )
{
Scanner input = new Scanner( System.in );
System.out.print( "Would you like to convert English to Binary (yes or no)? " );
String answer = input.nextLine();
if( answer.equals( "yes" ) )
{
System.out.println( "Enter English text here: " );
String english = input.nextLine();
System.out.println( stringToBinary( english ) );
}
if (answer.equalsIgnoreCase( "no" ) )
{
System.out.print( "Binary to English? " );
String answer2 = input.nextLine();
if (answer2.equalsIgnoreCase( "yes" ) )
{
System.out.println( "Enter Binary text here: " );
// enter binary here
String code = input.nextLine();
System.out.println( stringToEnglish( code ) );
}
}
}
public static String encode (String toEncode)
{
String binary = toEncode;
if (toEncode.equalsIgnoreCase("a"))
binary = "00001";
if (toEncode.equalsIgnoreCase("b"))
binary = "00010";
if (toEncode.equalsIgnoreCase("c"))
binary = "00011";
if (toEncode.equalsIgnoreCase("d"))
binary = "00100";
if (toEncode.equalsIgnoreCase("e"))
binary = "00101";
if (toEncode.equalsIgnoreCase("f"))
binary = "00110";
if (toEncode.equalsIgnoreCase("g"))
binary = "00111";
if (toEncode.equalsIgnoreCase("h"))
binary = "01000";
if (toEncode.equalsIgnoreCase("i"))
binary = "01001";
if (toEncode.equalsIgnoreCase("j"))
binary = "01010";
if (toEncode.equalsIgnoreCase("k"))
binary = "01011";
if (toEncode.equalsIgnoreCase("l"))
binary = "01100";
if (toEncode.equalsIgnoreCase("m"))
binary = "01101";
if (toEncode.equalsIgnoreCase("n"))
binary = "01110";
if (toEncode.equalsIgnoreCase("o"))
binary = "01111";
if (toEncode.equalsIgnoreCase("p"))
binary = "10000";
if (toEncode.equalsIgnoreCase("q"))
binary = "10001";
if (toEncode.equalsIgnoreCase("r"))
binary = "10010";
if (toEncode.equalsIgnoreCase("s"))
binary = "10011";
if (toEncode.equalsIgnoreCase("t"))
binary = "10100";
if (toEncode.equalsIgnoreCase("u"))
binary = "10101";
if (toEncode.equalsIgnoreCase("v"))
binary = "10110";
if (toEncode.equalsIgnoreCase("w"))
binary = "10111";
if (toEncode.equalsIgnoreCase("x"))
binary = "11000";
if (toEncode.equalsIgnoreCase("y"))
binary = "11001";
if (toEncode.equalsIgnoreCase("z"))
binary = "11010";
if (toEncode.equalsIgnoreCase(""))
binary = " ";
return binary;
}
public static String stringToBinary( String text )
{
String newText = "";
String selectedChar;
String convertedChar;
for (int i = 0; i < text.length(); i++)
{
selectedChar = text.charAt(i) + "";
convertedChar = encode(selectedChar);
if (convertedChar.equals(" "))
{
newText = newText + " ";
}
else
{
newText = newText + convertedChar;
if (!convertedChar.equals(" "))
{
newText = newText + " ";
}
}
}
return newText;
}
public static String stringToEnglish( String text )
{
String english = text;
if (text.equalsIgnoreCase("00001"))
english = "a";
if (text.equalsIgnoreCase("00010"))
english = "b";
if (text.equalsIgnoreCase("00011"))
english = "c";
if (text.equalsIgnoreCase("00100"))
english = "d";
if (text.equalsIgnoreCase("00101"))
english = "e";
if (text.equalsIgnoreCase("00110"))
english = "f";
if (text.equalsIgnoreCase("00111"))
english = "g";
if (text.equalsIgnoreCase("01000"))
english = "h";
if (text.equalsIgnoreCase("01001"))
english = "i";
if (text.equalsIgnoreCase("01010"))
english = "j";
if (text.equalsIgnoreCase("01011"))
english = "k";
if (text.equalsIgnoreCase("01100"))
english = "l";
if (text.equalsIgnoreCase("01101"))
english = "m";
if (text.equalsIgnoreCase("01110"))
english = "n";
if (text.equalsIgnoreCase("01111"))
english = "o";
if (text.equalsIgnoreCase("10000"))
english = "p";
if (text.equalsIgnoreCase("10001"))
english = "q";
if (text.equalsIgnoreCase("10010"))
english = "r";
if (text.equalsIgnoreCase("10011"))
english = "s";
if (text.equalsIgnoreCase("10100"))
english = "t";
if (text.equalsIgnoreCase("10101"))
english = "u";
if (text.equalsIgnoreCase("10110"))
english = "v";
if (text.equalsIgnoreCase("10111"))
english = "w";
if (text.equalsIgnoreCase("11000"))
english = "x";
if (text.equalsIgnoreCase("11001"))
english = "y";
if (text.equalsIgnoreCase("11010"))
english = "z";
if (text.equalsIgnoreCase(" "))
english = " ";
return english ;
}
}
答案 0 :(得分：3)
您的stringToEnglish方法只计算单个字符，因此它永远不会返回有效结果。你需要的是一个循环，StringBuilder以及比较中的一些变化。
首先，使用StringBuilder是构建字符串的最佳选择。字符串是不可变的，因此实际上每次添加一个时都会返回一个新实例。 StringBuilder是解决这个问题的方法。
public static String stringToEnglish( String text )
{
StringBuilder builder = new StringBuilder();
...
我们还需要注意您有两个实例需要处理：
所以你需要处理多个字符的循环看起来像......
for (int i = 0; i < text.length() - 4; ++i) {
...
优化是在它的二进制时跳转到字符串的末尾。 我们还需要考虑通过查看子字符串来迭代原始字符串......
if (text.charAt(i) == ' ') {
continue;
} else if (text.substring(i, i + 5).equalsIgnoreCase("00001")){
builder.append( "a" );
}
...
首先，我们会看看它是否以" "开头，如果是，请继续。使用else if而不是ifs链来避免不必要的比较。如果没有，我们知道它是一个二进制表示，需要将其作为子字符串拉出来并根据你的表格进行测试＆＃34 ;;当你找到它时 - 追加它。
我们一直持续到我们能够走到尽头，然后我们将返回我们已建成的......
return builder.toString() ;
}
您还可以switch使用break，或者更好地在评论中使用我的建议，而不是使用long if链。