本文介绍了java caesar密码的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
这是我的代码
public class CaesarCipher
{
public static final String ALPHABET =abcdefghijklmnopqrstuvwxyz;
public static String encrypt（String plainText，int shiftKey）
{
plainText = plainText.toLowerCase（）;
String cipherText =; （int i = 0; i< plainText.length（）; i ++）
{
int charPosition = ALPHABET.indexOf（plainText.charAt（i））;
int keyVal =（shiftKey + charPosition）％26;
char replaceVal = ALPHABET.charAt（keyVal）;
cipherText + = replaceVal;
}
return cipherText;
}
public static String decrypt（String cipherText，int shiftKey）
{
cipherText = cipherText.toLowerCase（）;
String plainText =; （int i = 0; i< cipherText.length（）; i ++）
{
int charPosition = ALPHABET.indexOf（cipherText.charAt（i））;
int keyVal =（charPosition - shiftKey）％26;
if（keyVal< 0）
{
keyVal = ALPHABET.length（）+ keyVal;
}
char replaceVal = ALPHABET.charAt（keyVal）;
plainText + = replaceVal;
}
返回plainText;
}
public static void main（String [] args）
{
扫描仪sc =新扫描仪（System.in）;
System.out.println（输入加密字符串）;
String message = new String（）;
message = sc.next（）;
System.out.println（encrypt（message，3））;
System.out.println（decrypt（encrypt（message，3），3））;
sc.close（）;
}
}
运行：
输入纯文本：
Reem LA
输入密钥：
2
密码文本
解决方案
使用 indexOf 不是非常有效...您可以在 char 值上执行整数运算以获取其索引。
我在代码中添加了注释来解释更多，但这是我想出来的。
public class CaesarCipher {
//旋转一个字符k位置
public static char cipher c，int k）{
//声明一些帮助常量
final int alphaLength = 26;
final char asciiShift = Character.isUpperCase（c）？ 'A'：'a'
final int cipherShift = k％alphaLength;
//转移到0..25为a..z
char shif =（char）（c - asciiShift）;
//旋转字母并处理负面和值的wrap-around> = 26
shif =（char）（（shift + cipherShift + alphaLength）％alphaLength）;
//换回英文字符
return（char）（shift + asciiShift）;
}
//旋转一个字符串k位置
public static String cipher（String s，int k）{
StringBuilder sb = new StringBuilder（）; （int i = 0; i< s.length（）; i ++）{
sb.append（cipher（s.charAt（i），k））;
}
return sb.toString（）;
}
public static void main（String [] args）{
扫描仪键盘=新的扫描仪（System.in）;
字符串密码;
int key;
System.out.print（请输入密码：）;
password = keyboard.nextLine（）;
do {
System.out.print（请输入1-25之间的键：）;
key = keyboard.nextInt（）;
if（key< 1 || key> 25）{
System.out.printf（该键必须介于1和25之间，您输入％d.\\\
，键）;
}
} while（key< 1 || key> 25）;
System.out.println（密码：\t+密码）;
字符串加密=密码（密码，密钥）;
System.out.println（加密：\t+加密）;
System.out.println（Decrypted：\t+ cipher（encryption，-key））;
}
}
输出应该是像/ p>
请输入密码：ABCDEFGHIJKLMNOPQRSTUVWXYZ
请输入1-25：1之间的密钥
密码：ABCDEFGHIJKLMNOPQRSTUVWXYZ
加密：BCDEFGHIJKLMNOPQRSTUVWXYZA
解密：ABCDEFGHIJKLMNOPQRSTUVWXYZ
i did caesar cipher code by java it runs but doesnt encrypt anything after user enter the key !
here is my code
public class CaesarCipher
{
public static final String ALPHABET = "abcdefghijklmnopqrstuvwxyz";
public static String encrypt(String plainText, int shiftKey)
{
plainText = plainText.toLowerCase();
String cipherText = "";
for (int i = 0; i < plainText.length(); i++)
{
int charPosition = ALPHABET.indexOf(plainText.charAt(i));
int keyVal = (shiftKey + charPosition) % 26;
char replaceVal = ALPHABET.charAt(keyVal);
cipherText += replaceVal;
}
return cipherText;
}
public static String decrypt(String cipherText, int shiftKey)
{
cipherText = cipherText.toLowerCase();
String plainText = "";
for (int i = 0; i < cipherText.length(); i++)
{
int charPosition = ALPHABET.indexOf(cipherText.charAt(i));
int keyVal = (charPosition - shiftKey) % 26;
if (keyVal < 0)
{
keyVal = ALPHABET.length() + keyVal;
}
char replaceVal = ALPHABET.charAt(keyVal);
plainText += replaceVal;
}
return plainText;
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
System.out.println("Enter the String for Encryption: ");
String message = new String();
message = sc.next();
System.out.println(encrypt(message, 3));
System.out.println(decrypt(encrypt(message, 3), 3));
sc.close();
}
}
run:
Enter The Plain Text:
Reem LA
Enter The Key:
2
The Cipher Text
解决方案
Using indexOf is not very efficient... You can do integer arithmetic on char values to get their indices.
I included comments in the code to explain more, but this is what I came up with.
public class CaesarCipher {
// Rotate a character k-positions
public static char cipher(char c, int k) {
// declare some helping constants
final int alphaLength = 26;
final char asciiShift = Character.isUpperCase(c) ? 'A' : 'a';
final int cipherShift = k % alphaLength;
// shift down to 0..25 for a..z
char shifted = (char) (c - asciiShift);
// rotate the letter and handle "wrap-around" for negatives and value >= 26
shifted = (char) ((shifted + cipherShift + alphaLength) % alphaLength);
// shift back up to english characters
return (char) (shifted + asciiShift);
}
// Rotate a string k-positions
public static String cipher(String s, int k) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < s.length(); i++) {
sb.append(cipher(s.charAt(i), k));
}
return sb.toString();
}
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
String password;
int key;
System.out.print("Please enter a password: ");
password = keyboard.nextLine();
do {
System.out.print("Please enter a key between 1-25: ");
key = keyboard.nextInt();
if (key < 1 || key > 25) {
System.out.printf(" The key must be between 1 and 25, you entered %d.\n", key);
}
} while (key < 1 || key > 25);
System.out.println("Password:\t" + password);
String encryption = cipher(password, key);
System.out.println("Encrypted:\t" + encryption);
System.out.println("Decrypted:\t" + cipher(encryption, -key));
}
}
The output should be something like
Please enter a password: ABCDEFGHIJKLMNOPQRSTUVWXYZ
Please enter a key between 1-25: 1
Password: ABCDEFGHIJKLMNOPQRSTUVWXYZ
Encrypted: BCDEFGHIJKLMNOPQRSTUVWXYZA
Decrypted: ABCDEFGHIJKLMNOPQRSTUVWXYZ
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