我对通过的V @ genere在Java中加密的字符串这个简单的功能。 我省略了解密，因为这仅仅是一个“ - ”，而不是在该行“+”，在那里计算的新值。
不过这个功能仅适用于正常的字母AZ。 我怎样才能改变功能，所以它支持小写字母以及大写字母和所有其他UTF-8字符？
public static String vigenere_encrypt(String plaintext, String key) {
String encryptedText = "";
for (int i = 0, j = 0; i < plaintext.length(); i++, j++) {
if (j == key.length()) { j = 0; } // use key again if end reached
encryptedText += (char) ((plaintext.charAt(i)+key.charAt(j)-130)%26 + 65);
}
return encryptedText;
}
非常感谢您的帮助！
另一种答案，那确实做上限与小写字符的V @ genere加密，只需插入其他字符。 使用这种技术来创建字符编码的多个组。
public static String vigenere(String plaintext, String key, boolean encrypt) {
final int textSize = plaintext.length();
final int keySize = key.length();
final int groupSize1 = 'Z' - 'A' + 1;
final int groupSize2 = 'z' - 'a' + 1;
final int totalGroupSize = groupSize1 + groupSize2;
final StringBuilder encryptedText = new StringBuilder(textSize);
for (int i = 0; i < textSize; i++) {
final char plainChar = plaintext.charAt(i);
// this should be a method, called for both the plain text as well as the key
final int plainGroupNumber;
if (plainChar >= 'A' && plainChar <= 'Z') {
plainGroupNumber = plainChar - 'A';
} else if (plainChar >= 'a' && plainChar <= 'z') {
plainGroupNumber = groupSize1 + plainChar - 'a';
} else {
// simply leave spaces and other characters
encryptedText.append(plainChar);
continue;
}
final char keyChar = key.charAt(i % keySize);
final int keyGroupNumber;
if (keyChar >= 'A' && keyChar <= 'Z') {
keyGroupNumber = keyChar - 'A';
} else if (keyChar >= 'a' && keyChar <= 'z') {
keyGroupNumber = groupSize1 + keyChar - 'a';
} else {
throw new IllegalStateException("Invalid character in key");
}
// this should be a separate method
final int cipherGroupNumber;
if (encrypt) {
cipherGroupNumber = (plainGroupNumber + keyGroupNumber) % totalGroupSize;
} else {
// some code to go around the awkward way of handling % in Java for negative numbers
final int someCipherGroupNumber = plainGroupNumber - keyGroupNumber;
if (someCipherGroupNumber < 0) {
cipherGroupNumber = (someCipherGroupNumber + totalGroupSize);
} else {
cipherGroupNumber = someCipherGroupNumber;
}
}
// this should be a separate method
final char cipherChar;
if (cipherGroupNumber < groupSize1) {
cipherChar = (char) ('A' + cipherGroupNumber);
} else {
cipherChar = (char) ('a' + cipherGroupNumber - groupSize1);
}
encryptedText.append(cipherChar);
}
return encryptedText.toString();
}
再次，这是不安全的代码所使用的密码已被打破的年龄。 不要在你的钥匙使用太多的“A”字:)但字符编码应该是合理的。
嗯，你问它，我觉得有些莫名其妙，但打印出来的密文，你就会知道你刚才问的...
public static String vigenereUNICODE(String plaintext, String key, boolean encrypt) {
final int textSize = plaintext.length();
final int keySize = key.length();
final StringBuilder encryptedText = new StringBuilder(textSize);
for (int i = 0; i < textSize; i++) {
final int plainNR = plaintext.codePointAt(i);
final int keyNR = key.codePointAt(i % keySize);
final long cipherNR;
if (encrypt) {
cipherNR = ((long) plainNR + (long) keyNR) & 0xFFFFFFFFL;
} else {
cipherNR = ((long) plainNR - (long) keyNR) & 0xFFFFFFFFL;
}
encryptedText.appendCodePoint((int) cipherNR);
}
return encryptedText.toString();
}
编辑：请永远不要在生产代码中使用，因为我还没有得到，如果代码点的确可以编码/解码的线索。 并非所有的点已经定义，因为据我所知，标准是一个移动的目标。
如果完整的Unicode支持是不可能的，你必须确定你的有效字符，列表反正，为什么不使用像这样的功能？
public static String vigenere_cipher(String plaintext, String key, boolean encrypt) {
String alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ ,.-"; // including some special chars
final int alphabetSize = alphabet.length();
final int textSize = plaintext.length();
final int keySize = key.length();
final StringBuilder encryptedText = new StringBuilder(textSize);
for (int i = 0; i < textSize; i++) {
final char plainChar = plaintext.charAt(i); // get the current character to be shifted
final char keyChar = key.charAt(i % keySize); // use key again if the end is reached
final int plainPos = alphabet.indexOf(plainChar); // plain character's position in alphabet string
if (plainPos == -1) { // if character not in alphabet just append unshifted one to the result text
encryptedText.append(plainChar);
}
else { // if character is in alphabet shift it and append the new character to the result text
final int keyPos = alphabet.indexOf(keyChar); // key character's position in alphabet string
if (encrypt) { // encrypt the input text
encryptedText.append(alphabet.charAt((plainPos+keyPos) % alphabetSize));
}
else { // decrypt the input text
int shiftedPos = plainPos-keyPos;
if (shiftedPos < 0) { // negative numbers cannot be handled with modulo
shiftedPos += alphabetSize;
}
encryptedText.append(alphabet.charAt(shiftedPos));
}
}
}
return encryptedText.toString();
}
这应该是一个很短，工作版本。 和字母表可以容易地存储在可总是被扩展的字符串（其导致不同的密文）。
