我最初遇到了一些问题,但在这里找到了一些不同的帮助。现在我似乎遇到输入异常错误的问题。我相信我的输入格式正确。
import java.util.Scanner;
public class CaesarShift
{
//initialize private string for the alphabet
private final String ALPHABET = "abcdefghijklmnopqrstuvwxyz";
//public encryption code
public String encryptionMethod(String normText, int caesarShift)
{
normText = normText.toLowerCase();
String cipherText = "";
for (int a = 0; a < normText.length(); a++)
{
int charP = ALPHABET.indexOf(normText.charAt(a));
int shiftValue = (caesarShift + charP) % 26;
char replaceValue = this.ALPHABET.charAt(shiftValue);
cipherText += replaceValue;
}
return cipherText;
}
public String decryptionMethod(String cipherText,int caesarShift)
{
cipherText = cipherText.toLowerCase();
String normText = "";
for (int a = 0; a < cipherText.length(); a++)
{
int charP = this.ALPHABET.indexOf(cipherText.charAt(a));
int keyValue = (charP - caesarShift) % 26;
if(keyValue < 0)
{
keyValue = this.ALPHABET.length() + keyValue;
}
char replaceValue = this.ALPHABET.charAt(keyValue);
normText += replaceValue;
}
return normText;
}
}
然后我有测试器方法,我遇到输入异常错误的实际问题
import java.util.Scanner;
public class CaesarShiftTester
{
public static void main(String args[])
{
//import of the scanner method to ask the user for the input they would like
Scanner in = new Scanner(System.in);
System.out.println("What is the text you would like to do something with?");
String normText = in.next();
System.out.println("What is the Caesar Shift Value?");
int caesarShift = in.nextInt();
//new declaration of the CaesarShift class to report back to easily
CaesarShift shift = new CaesarShift();
//decalre the need properties for the encryption
String cipherText = shift.encryptionMethod(normText, caesarShift);
System.out.println("Your normal text is: " + normText);
System.out.println("Your text after encryption is: " + cipherText);
String cnormText = shift.decryptionMethod(cipherText, caesarShift);
System.out.println("Your encrypted text is: " + cipherText);
System.out.println("Your decrypte text is: " + cnormText);
}
}
对于有些混乱的代码感到抱歉,我通常会在程序完成并正常工作时进行清理。
答案 0 :(得分:1)
如果您只输入1个单词,您的程序应该可以正常工作。如果包含空格,则会出现异常。问题在线
String normText = in.next();
应该是
String normText = in.nextLine();
将整行作为输入text
。 next()
无效,因为
Scanner
使用分隔符模式将其输入分解为标记, 默认情况下匹配空格。
因此,它仅匹配第一个单词,并尝试将下一个单词解析为int
(因为您的下一行int caesarShift = in.nextInt();
)
其他一些观点:
char
是否是一个字母(例如使用Character.isLetter()
)并仅移动那些字符(目前,它不会在{{ 1}}所以ALPHABET
返回indexOf
)-1
,它更快