Question

我遇到的问题是用凯撒密码程序包裹字母表。

该程序适用于所有小写字母。完美包裹，能够应用正面和负面的转变。当我尝试输入一个大写字母时，大写字母不会回绕。

这是我的代码：

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();
}

}

提示非常感谢！当然，我不是要求任何人为我做我的工作，但一些帮助将不胜感激！谢谢！：）

编辑：这是仅包含小写字母的if语句：

if(strChars[i] > 'z') 
{
 strChars[i] -= 'z';
 strChars[i] += ('a' - 1);         
}

Answer 1

让我们为单个字符实现环绕功能。这将由另一种方法使用。当您明智地分离任务和子任务时，您会发现问题变得更容易解决。在这里，我的解决方案基于<nav class="navbar navbar-default navbar-fixed-top" id="barrademenu"> <div class="container"> <div class="navbar-header"> <a class="navbar-brand" href="#">Logo</a> </div> <div class="collapse navbar-collapse" id="myNavbar"> <ul class="nav navbar-nav navbar-right"> <li><a href="#home" id="home">HOME</a></li> <li><a href="#about" id="about">ABOUT US</a></li> <li><a href="#portfolio" id="portfolio">PORTFOLIO</a></li> <li><a href="#contact" id="contact">CONTACT US</a></li> </ul> </div> </div> </nav>变量表示为数字的事实，我们知道字母的数量，我们可以将该数字用作模数类，以确保代数帮助我们。

char

现在，而不是

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));
}

你需要：

    //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);         
        }**

凯撒密码大写和小写环绕

1 个答案: