Java中的Caesar密码（西班牙语字符）

Question

我已经阅读了这个问题 ，我想知道是否有任何方法可以考虑整个字符范围？ 例如，“á”，“é”，“ö”，“ñ”，而不考虑“”（[空格]）？ （例如，我的String是“ Hello World”，标准结果是“ Khoor＃Zruog”；我想删除该“＃”，因此结果将是“ KhoorZruog”）

我确定我的答案在这段代码中：

if (c >= 32 && c <= 127)
        {
            // Change base to make life easier, and use an
            // int explicitly to avoid worrying... cast later
            int x = c - 32;
            x = (x + shift) % 96;
            chars[i] = (char) (x + 32);
        }

但是我已经尝试了一些方法，但是没有用。

Answer 1

参见以下伪代码-应该可以轻松实现：

// you need to define your own range, obviously - it's not at all obvious whether
// e.g. "ź" should be included and that it should come after "z"
array char_range = ['a','á','b','c','č', (...), 'z','ź','ž'] 
// the text to encode
string plaintext = 'some text here'
// this will contain encoded text
stringbuilder ciphertext = ''
// the classic Caesar Cipher shifts by 3 chars to the right
// to decipher, reverse the sign
int shift_by = 3 
// note: character != byte, esp. not in UTF-8 (1 char could be 1 or more bytes)
for each character in plaintext
    get character_position of character in char_range // e.g. "a" would return 0
    if not in char_range // e.g. spaces and other non-letters
        do nothing // drop character 
        // alternately, you can append it to ciphertext unmodified
        continue with next character
    add shift_by to character_position
    if character_position > char_range.length
        character_position modulo char_range.length
    if character_position < 0 // useful for decoding
        add char_range.length to character_position 
    get new_character at character_position
    append new_character to ciphertext
done

Answer 2

不过滤的ASCII码32空格。 您可以尝试：

if (c >= 33 && c <= 127) 
    { 
        // Change base to make life easier, and use an 
        // int explicitly to avoid worrying... cast later 
        int x = c - 32; 
        x = (x + shift) % 96; 
        chars[i] = (char) (x + 32); 
    } 

我只是在if-Clause中用32更改了32，以便简单地忽略空格。

Answer 3

您可以使用它。 它将检查您是否给定的int值表示文字。 字符

因此您的函数可能如下所示：

if (Character.isLiteral(c) )
{
     // Change base to make life easier, and use an
     // int explicitly to avoid worrying... cast later
     int x = c - Character.MIN_VALUE;
     x = (x + shift) % Character.MAX_VALUE;
     chars[i] = (char) (x + Character.MIN_VALUE);
}