java凯撒密码 [英] java caesar cipher code

问题描述

我用它运行的java做了凯撒密码，但在用户输入密钥后不加密任何东西！

i did caesar cipher code by java it runs but doesnt encrypt anything after user enter the key !

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 = "";
        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();
    }
}

运行:

Enter The Plain Text:
Reem LA
Enter The Key:
2
The Cipher Text

推荐答案

使用 indexOf 不是很有效...你可以对 char 值做整数运算得到他们的指数.

Using indexOf is not very efficient... You can do integer arithmetic on char values to get their indices.

我在代码中加入了注释来解释更多，但这就是我想出来的.

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.
", key);
            }
        } while (key < 1 || key > 25);


        System.out.println("Password:	" + password);
        String encryption = cipher(password, key);
        System.out.println("Encrypted:	" + encryption);
        System.out.println("Decrypted:	" + cipher(encryption, -key));

    }
}

输出应该类似于

Please enter a password: ABCDEFGHIJKLMNOPQRSTUVWXYZ
Please enter a key between 1-25: 1
Password:   ABCDEFGHIJKLMNOPQRSTUVWXYZ
Encrypted:  BCDEFGHIJKLMNOPQRSTUVWXYZA
Decrypted:  ABCDEFGHIJKLMNOPQRSTUVWXYZ

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