java - Java 中的凯撒密码(西类牙语字符)

Question

我已阅读this question ，我想知道是否有办法考虑整个字符范围？例如“á”、“é”、“ö”、“ñ”，而不考虑“”([空格])？ (例如，我的字符串是“Hello World”，标准结果是“Khoor#Zruog”；我想删除那个“#”，所以结果将是“KhoorZruog”)

我确信我的答案就在这段代码中:

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);
        }

但是我尝试了一些方法，但没有成功。

Answer 1

请参阅此伪代码 - 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