凯撒密码移位(使用字母数组)

我对您的代码进行了一些处理。下面给出了解决方案，但你必须小心：你只能使用大写字母，因为上下图表有区别。我使用了ToUpper（）方法。对我来说很好。我想这就是你的问题所在。

public static void Main()
    {
        string encrypted_text = "BCD";     //String variable that contains the text from a file. To get the text, the method in a class SystemIO is ran to read the text. It expects a parameter, which is a file directory.
        string decoded_text = " ";
        int shift = 0;
        char character = '0';
        encrypted_text = encrypted_text.ToUpper();
        char[] alphabet = new char[26] { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z' };
        Console.WriteLine("The encrypted text is 'n{0}", encrypted_text);       //Display the encrypted text
        for (int i = 0; i < alphabet.Length; i++)        //Start a loop which will display 25 different candidates of decipher
        {
            decoded_text = "";
            foreach (char c in encrypted_text)
            {
                character = c;
                if (character == '''' || character == ' ')
                    continue;
                shift = Array.IndexOf(alphabet, character) - i;     //Define a shift which is the index of a character in an alphabet array, take away the itteration of this loop. Store the result in a variable
                if (shift <= 0)
                    shift = shift + 26;
                if (shift >= 26)
                    shift = shift - 26;

                decoded_text += alphabet[shift];
            }
            Console.WriteLine("'nShift {0} 'n {1}", i + 1, decoded_text);
        }
    }

我查看了您的代码，并进行了轻微调整。首先，我将其转换为一个方法，允许您传入字符串和要偏移的量，这样您就可以在从0到25的循环中调用它来查看所有排列，也可以只获得一个值。我还检查每个字符是否真的在数组中，如果不是，那么不要更改它（在代码中，您只检查'''和' '字符：

public static string ShiftText(string input, int shiftAmount)
{
    if (input == null) return null;
    char[] alphabet =
    {
        'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M',
        'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'
    };
    shiftAmount %= 26; // Ensure shift is between 0 and 25
    var shiftedText = string.Empty;
    foreach (var character in input)
    {
        var index = Array.IndexOf(alphabet, character);
        if (index < 0)
        {
            // This character isn't in the array, so don't change it
            shiftedText += character;
        }
        else
        {
            var newIndex = index - shiftAmount;
            // If it's negative, wrap around to end of array
            if (newIndex < 0) newIndex += 26;
            shiftedText += alphabet[newIndex];
        }
    }
    return shiftedText;
}

但是，另一种适用于大写和小写并且代码较少的方法是简单地测试是否为char.IsLetter(character)，然后在相同的0-25范围内移动字符的ASCII值。

例如，这与上面的代码相同，只是它也适用于小写字母。这里的区别在于，在我们将该字符与我们的最低值字符（'a'或'A'）进行比较之前，我们首先测试是否为char.IsLower()。这样，我们就可以保持在这个字符集的ASCII范围内：

/// <summary>
/// This method takes the input string and shifts all letter characters 
/// to the left (subtracts) by the amount specified in shiftAmount, so 
/// if shiftAmount = 1, then 'M' becomes 'L', and 'a' becomes 'z'.
/// </summary>
/// <param name="input">The input string to apply changes to</param>
/// <param name="shiftAmount">A value from 0 to 25, used to shift the characters</param>
/// <returns>The modified (shifted) string</returns>
public static string ShiftText(string input, int shiftAmount)
{
    if (input == null) return null;
    // Ensure shift is between 0 and 25
    shiftAmount %= 26; 
    var result = string.Empty;
    // Loop through input and update result with shifted letters
    foreach (var character in input)
    {
        if (!char.IsLetter(character))
        {
            // If the character isn't a letter, don't change it
            result += character;
        }
        else
        {
            var newChar = (char) (character - shiftAmount);
            // Adjust newChar to stay within this character range
            if (newChar < (char.IsLower(character) ? 'a' : 'A')) newChar += (char) 26;
            result += newChar;
        }
    }
    return result;
}

为什么不直接使用字符的ASCII值呢。我会先把密文转换成小写。例如，a的asci值为97。我会写一个方法来提取97个每个字符，所以a=0，b=1……z=25。然后，对于密文中的每个字符，得到该字符的-3偏移值。例如，输入字符d应返回值0，该值对应于a。