我需要为一个赋值编写一个简单的Caesar密码,我必须加密消息“This is a Caesar cipher”,左移为3.我尝试使用IF语句后跟'continue;'但它不起作用,我不能为我的生活找出导致这个问题的原因哈哈。
public static String encrypt(String plainText, int shiftKey) {
plainText = plainText.toLowerCase();
String cipherText = "";
for (int i = 0; i < plainText.length(); i++) {
char replaceVal = plainText.charAt(i);
int charPosition = ALPHABET.indexOf(replaceVal);
if(charPosition != -1) {
int keyVal = (shiftKey + charPosition) % 26;
replaceVal = ALPHABET.charAt(keyVal);
}
cipherText += replaceVal;
}
return cipherText;
}
public static void main (String[] args) {
String message;
try (Scanner sc = new Scanner(System.in)) {
System.out.println("Enter a sentence to be encrypted");
message = new String();
message = sc.next();
}
System.out.println("The encrypted message is");
System.out.println(encrypt(message, 23));
}
}
你只用Scanner.next()
读一个单词,从不使用new String()
。更改
message = new String();
message = sc.next();
至
message = sc.nextLine();
值得注意的是,StringBuilder
和简单算术就是凯撒密码所需要的。例如,
public static String encrypt(String plainText, int shiftKey) {
StringBuilder sb = new StringBuilder(plainText);
for (int i = 0; i < sb.length(); i++) {
char ch = sb.charAt(i);
if (!Character.isWhitespace(ch)) {
sb.setCharAt(i, (char) (ch + shiftKey));
}
}
return sb.toString();
}
public static void main(String[] args) {
int key = 10;
String enc = encrypt("Secret Messages Are Fun!", key);
System.out.println(enc);
System.out.println(encrypt(enc, -key));
}
哪个输出
]om|o~ Wo}}kqo} K|o Px+
Secret Messages Are Fun!