Java Caesar密码大小写环绕

我有一个问题，关于用我的凯撒密码程序包装字母表

该程序适用于所有小写字母。完美环绕，能够应用正、负移位。当我尝试输入大写字母时，大写字母不会环绕

这是我的密码：

public static StringBuilder encode(String str, int numShift)
{

    numShift = numShift % 26;

    //assign each character of the string to a position in the character array
    char[] strChars = str.toCharArray();

    for (int i = 0; i < strChars.length; i++)
        {
            //ignore and continue if the character is not within the alphabet
            if((strChars[i] < 'a' || strChars[i] > 'z') && (strChars[i]<'A' || strChars[i] > 'Z')) 
                continue;

            //apply the shift to each character
            strChars[i] += numShift;

            //wrap around if the shift is beyond Z
            **if(strChars[i] > 'z') 
                {
                strChars[i] -= 'z';
                strChars[i] += ('a' - 1);         
                }**

        }

    StringBuilder encodedStr = new StringBuilder();

    encodedStr.append(strChars);

    return encodedStr;


}
public static void init(){
    Scanner scan = new Scanner(System.in);
    System.out.println("Please enter the string that you would like to  encode:");
    String str = scan.nextLine();
    System.out.println("Please enter the number of letters you would like to shift:");
    int strShift = scan.nextInt();
    scan.close();

    StringBuilder result = encode(str, strShift);
    System.out.println(result);
}
public static void main(String[] args)
{
        init();
}

---

让我们为单个字符实现一个环绕函数。这将由另一种方法使用。当你明智地将任务和子任务分开时，你会发现你的问题变得更容易解决。在这里，我的解决方案基于这样一个事实，即

char

变量被表示为数字，我们知道字母的数量，我们可以使用这个数字作为模类，以确保代数对我们有帮助

private static char wrapChar(char input, int amount) {
    //assume for now that we have an upper-case letter
    char start = 'A';
    //if the assumption is mistaken...
    if (('a' <= input) && (input <= 'z')) {
        //then, if it is lower-case, then use lower-case
        start = 'a';
    } else if (!(('A' <= input) && (input <= 'Z'))) {
        //target is not letter
        return input;
    }
    //Calculate total offset compared to the first letter of the alphabet
    //be it lower or upper
    int offset = ((input - start) + amount) % 26;
    //If offset happens to be negative, then shift a modulo period of 26
    //To get the correct positive offset
    if (offset < 26) {
        offset += 26;
    }
    //Add the final offset to start and convert it to char
    return ((char)(start + offset));
}

你只需要：

strChars[i] = wrapChar(strChars[i], numShift);

---

您的方法无法按现状工作，因为+7或更大的移位将使一些大写字母移到小写字母的范围内。我建议在转换前检查字母是否在大写范围内。如果是，请将结果转换为小写、转换并转换回。否则，就像你已经做的那样转变。这是有道理的，谢谢你，约翰！

    //apply the shift to each character
    strChars[i] += numShift;

    //wrap around if the shift is beyond Z
    **if(strChars[i] > 'z') 
        {
        strChars[i] -= 'z';
        strChars[i] += ('a' - 1);         
        }**

strChars[i] = wrapChar(strChars[i], numShift);