正则表达式删除空格和空行

Question

我正在尝试创建一个将文本文件解密为可读文本的程序。编码文本遵循两个规则，它总是以元音或辅音开头，表示为 C 或 V，后跟其索引位置（例如：V1C1C3，将是“abc”）

我创建了两个数组来包含元音和辅音（“0”位于索引 0 处，因此数组的其余部分可以从索引 1 开始）：

String[] vowels = {"0", "a", "A", "e", "E", "i", "I", "o", "O", "u", "U", "y", "Y"};
String[] cons = {"0","b", "B", "c", "C", "d", "D", "f", "F", "g", "G", "h", "H", "j", "J", "k", "K", "l", "L", "m", "M", "n", "N", "p", "P", "q", "Q", "r", "R", "s", "S", "t", "T", "v", "V", "w", "W", "x", "X", "z", "Z"};

我正在使用扫描仪将代码分成元音和辅音，并获得索引值：

while(scan.hasNext()){
    String[] parts = scan.nextLine().split("(?=[CV])");

    for (String part : parts) {
        Scanner num = new Scanner(part).useDelimiter("[^0-9]+");
        int value = num.nextInt();

        if(part.charAt(0) == 'C'){
           System.out.print(cons[value]);
        }
        else if (part.charAt(0) == 'V'){
           System.out.print(vowels[value]);
        }
    }
}

密文：V6 C17V7C33V3 V1C23C23C17V3C29（结果应该是：我爱苹果）

我得到的结果：Iloveapples

P.S：如果我有多个加密文本段落，扫描仪会在第一段之后停止并输出错误。如果我使用“scan.next()”而不是“scan.nextLine()”，则不会发生这种情况

Answer 1

您似乎希望任何不是编码的部分保持不变，例如示例中的空格，但也可以是逗号、问号等符号。

这意味着使用正则表达式替换会更好，使用正则表达式来定位编码，即字母 C 或 V 后跟数字。

private static final String VOWELS = "aAeEiIoOuUyY";
private static final String CONSONANTS = "bBcCdDfFgGhHjJkKlLmMnNpPqQrRsStTvVwWxXzZ";

// Requires Java 9+
public static String decode(String encodedText) {
    return Pattern.compile("[CV]\\d+").matcher(encodedText)
            .replaceAll(r -> String.valueOf((r.group().charAt(0) == 'V' ? VOWELS : CONSONANTS)
                                            .charAt(Integer.parseInt(r.group().substring(1)) - 1)));
}

// For Java 1.4+
public static String decode(String encodedText) {
    StringBuffer buf = new StringBuffer();
    Matcher m = Pattern.compile("[CV]\\d+").matcher(encodedText);
    while (m.find()) {
        String token = m.group();
        int value = Integer.parseInt(token.substring(1));
        char ch = (token.charAt(0) == 'V' ? VOWELS : CONSONANTS).charAt(value - 1);
        m.appendReplacement(buf, String.valueOf(ch));
    }
    return m.appendTail(buf).toString();
}

测试

System.out.println(decode("V6 C17V7C33V3 V1C23C23C17V3C29"));
System.out.println(decode("C30C11V3 V1C29C15V3C5, \"C36C11V1C31 V1C27V3 V11V7V9 C5V7V5C21C9?\""));

输出

I love apples
She asked, "What are you doing?"

---

为了完整起见，这里是一个简单的未优化编码方法，用于创建上面的第二个示例：

// For Java 1.5+
public static String encode(String plainText) {
    int index;
    StringBuilder buf = new StringBuilder();
    for (int i = 0; i < plainText.length(); i++) {
        char ch = plainText.charAt(i);
        if ((index = VOWELS.indexOf(ch)) != -1)
            buf.append('V').append(index + 1);
        else if ((index = CONSONANTS.indexOf(ch)) != -1)
            buf.append('C').append(index + 1);
        else
            buf.append(ch);
    }
    return (buf.length() == plainText.length() ? plainText : buf.toString());
}

Answer 2

第一部分编码了一些测试短语。第二部分对它们进行解码，保留标点符号。

String[] vowels = { "0", "a", "A", "e", "E", "i", "I", "o",
        "O", "u", "U", "y", "Y" };
String[] cons = { "0", "b", "B", "c", "C", "d", "D", "f", "F",
        "g", "G", "h", "H", "j", "J", "k", "K", "l", "L", "m",
        "M", "n", "N", "p", "P", "q", "Q", "r", "R", "s", "S",
        "t", "T", "v", "V", "w", "W", "x", "X", "z", "Z" };

String c = String.join("", cons);
String v = String.join("", vowels);
    
String[] testData = { "I Love Apples\n",
        "To be or not to be that is the question!\n",
        "This also handles (parens) and `single` and \"double\" quotes\n" };
    
// this section iterates over the phrases and encodes them, storing the values in
// a StringBuilder
List<String> coded = new ArrayList<>();
for (String text : testData) {
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < text.length(); i++) {
        char ch = text.charAt(i);
        int index = v.indexOf(ch);
        if (index >= 0) {
            sb.append("V").append(index);
            continue;
        }
        index = c.indexOf(ch);
        if (index >= 0) {
            sb.append("C").append(index);
            continue;
        }
        sb.append(ch);
    }
    coded.add(sb.toString());
}
    
for (String code : coded) {
    System.out.println(code);
}
// Now simply search on encoded values or a single character. Keep finding the
// pattern and converting back to a letter.  Punctuation is preserved.
String pat = "([CV]\\d+|.)";
for (String code : coded) {
    StringBuilder plain = new StringBuilder();
    Matcher m = Pattern.compile(pat).matcher(code);
    while (m.find()) {
        String group = m.group();
        if (group.charAt(0) == 'V') {
            plain.append(v.charAt(
                    Integer.valueOf(group.substring(1))));
        } else if (group.charAt(0) == 'C') {
            plain.append(c.charAt(
                    Integer.valueOf(group.substring(1))));
        } else {
            plain.append(group);
        }
    }
    
    System.out.println(plain);
}

上面首先打印编码的字符串，然后是解码的字符串。

V6 C18V7C33V3 V2C23C23C17V3C29

C32V7 C1V3 V7C27 C21V7C31 C31V7 C1V3 C31C11V1C31 V5C29 C31C11V3 C25V9V3C29C31V5V
7C21!

C32C11V5C29 V1C17C29V7 C11V1C21C5C17V3C29 (C23V1C27V3C21C29) V1C21C5 `C29V5C21C9
C17V3` V1C21C5 "C5V7V9C1C17V3" C25V9V7C31V3C29

I Love Apples
To be or not to be that is the question!
This also handles (parens) and `single` and "double" quotes