在此程序方法中，encode和decode正在工作，但我必须包括第三种方法rotation。用户必须输入一个解码单词，例如"eadaf eals"，并且程序使用算法decode显示所有25个Possibiliies。
例如：
Rotation text
1 eadaf eal //original text from user
2 eolwk wkk
3 hello you
....
25 klsjd ddk
但是我对最后一种方法有问题。所有25个都必须有一个for循环可能性。
public class CaesarsChiffre {
public static void main(String[] args) {
char benutzerBefehl;
String benutzerText;
int rotation=0;
System.out.println("Caesars Chiffre encode, decode, rotate.");
System.out.println("Waehlen Sie:");
System.out.println("V for encode");
System.out.println("E for decode");
System.out.println("B for rotation");
benutzerBefehl=Input.readChar();
switch (benutzerBefehl){
case 'V': System.out.println("Enter text to encode:");
benutzerText=Input.readString();
System.out.println("which rotation do you want?");
rotation=Input.readInt();
while (rotation < 1 || rotation > 25) {
System.out.println(" The key must be between 1 and 25, you entered: "+ benutzerBefehl);
break;
}
System.out.print("Encoded text: ");
System.out.println(Caesarencode(benutzerText,rotation));
break;
case 'E': System.out.println("Enter text to decode:");
benutzerText=Input.readString();
System.out.println("which rotation do you want?");
rotation=Input.readInt();
while (rotation < 1 || rotation > 25) {
System.out.println(" The key must be between 1 and 25, you entered: "+ benutzerBefehl);
break;
}
System.out.println("Decoded Text :");
System.out.println(Caesardecode(benutzerText,rotation));
break;
case 'B': System.out.println("Enter the rotated text:");
benutzerText=Input.readString();
System.out.println("Rotation Text: ");
System.out.println(Caesardecode(benutzerText));
default:
System.out.println("Invalid option..");
}
}
public static String Caesarencode(String original_text, int rot){
int length =original_text.length();
String cypherText="";
for(int i=0;i<length;i++){
char ch=original_text.charAt(i);
if(Character.isLetter(ch)){
if(Character.isLowerCase(ch)){
char c=(char)(ch+rot);
if(c>'z'){
cypherText += (char)(ch-(26-rot));
}
else{
cypherText+=c;
}
}
else if(Character.isUpperCase(ch)){
char c=(char)(ch+rot);
if(c>'Z'){
cypherText += (char)(ch-(26-rot));
}
else{
cypherText += c;
}
}
}
else{
cypherText+=ch;
}
}
return cypherText;
}
public static String Caesardecode(String original_text, int rot){
int length =original_text.length();
String cypherText="";
for(int i=0;i<length;i++){
char ch=original_text.charAt(i);
if(Character.isLetter(ch)){
if(Character.isLowerCase(ch)){
char c=(char)(ch-rot);
if(c<'a'){
cypherText += (char)(ch+(26-rot));
}
else{
cypherText+=c;
}
}
else if(Character.isUpperCase(ch)){
char c=(char)(ch-rot);
if(c<'A'){
cypherText += (char)(ch+(26-rot));
}
else{
cypherText += c;
}
}
}
else{
cypherText+=ch;
}
}
return cypherText.toString();
}
public static String[] Caesardecode(String secret_text){
String[] textNeu= new String[25]; //problems with this part
for(int i=0;i<textNeu.length;i++){
textNeu[i]= Caesardecode(secret_text, i)+" ";
}
return textNeu;
}
}
对于您的问题，您正在打印一个数组
System.out.println(Caesardecode(benutzerText));
您可以看到
public static String[] Caesardecode(String secret_text){
您需要Arrays.toString(array)才能获得更可读的东西。
对于无效的选项是因为您忘记了case
break语句 case 'B': System.out.println("Enter the rotated text:");
benutzerText=Input.readString();
System.out.println("Rotation Text: ");
System.out.println(Caesardecode(benutzerText));
break; // ADD IT HERE
default:
System.out.println("Invalid option..");
}
仅仅因为您已经做了很多事情，这里是使用一种方法来增加参数的方法。
我有点清洁代码，使用了一些您错过的Character方法（较低/上情况。
但最重要的是，%26的使用允许z 1成为a
public static String ceasar(String s, int inc){
char[] input = s.toCharArray();
char[] output = new char[input.length];
for(int i = 0; i < output.length; ++i){
char c = input[i];
if(Character.isAlphabetic(c)){
char a;
if(Character.isUpperCase(c)){
a = 'A';
} else if(Character.isLowerCase(c)){
a = 'a';
} else { //not a letter, no evaluation needed
output[i] = input[i];
continue;
}
//encode the letter
output[i] = (char)(((c - a + inc) % 26) + a);
} else {
output[i] = input[i];
}
}
return new String(output);
}
您可以轻松地将其放入循环中
for(int i = 0; i < 26; ++i){
ceasar("youCode", i);
}