Question

我正在为我的大学项目创建一个简单的Java程序，但我无法制作数组字符串模数。该程序只需一个字符串，并从+5或-5加密或解密它，用于数组中的字母。但是如果我在输入字符串中输入数组中的最后5个中的任何一个（5,6,7,8,9）它返回什么，但它应该返回（a，b，c，d，e）开头的字母数组。

我必须使用的程序是非常原始的，只允许代码以某种方式，因此它的编码就像这样。

int i;
int n;
int x;
int inputValue;
int intArrayLength;
int intInputLength;
String array;
String inputString;
String newInputString;
String outputString;

System.out.print("Would you like to: 1 = Encrypt - 2 = Decrypt");
System.in.read(inputValue);
System.out.println("Would you like to: 1 = Encrypt - 2 = Decrypt [ " + inputValue + " ]");
System.out.print("Please enter string:");
System.in.read(inputString);
System.out.println("Please enter string: [ " + inputString + " ]");

array = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ 0123456789";
intArrayLength = length(array);
intInputLength = length(inputString);

if (intInputLength != 0)
   {
   if (inputValue == 1)
      {



      System.out.println("Please wait encrypting...");

      newInputString = "";
      for (i=0; i<=intInputLength; i++)
         {
         for (n=0; n<=intArrayLength; n++)
            {
            if (inputString[i] == array[n])
               {
               x = (n+5);
               if (x > intArrayLength)
                  {
                  x = x - intArrayLength;
                  newInputString = newInputString + array[x];
                  }
               else
                  {
                  newInputString = newInputString + array[x];
                  }
               }
            }
         }

      System.out.print("Would you like to reverse the string? 1 = Yes - 2 = No");
      System.in.read(inputValue);
      System.out.println("Would you like to reverse the string? 1 = Yes - 2 = No [ " + inputValue + " ]");

      if (inputValue == 1)
         {
         outputString = newInputString;
         newInputString = "";
         for (i=intInputLength-1; i>=0; i--)
            {
            newInputString = newInputString + outputString[i];
            }

         outputString = newInputString;
         System.out.print("Encrypted String: [ " + outputString + " ]");
         System.out.println();

         }
      else
         {
         outputString = newInputString;
         System.out.print("Encrypted String: [ " + outputString + " ]");
         System.out.println();
         }

      inputString = "";
      newInputString = "";



      }
   else
      {



      System.out.print("Has the string been reversed? 1 = Yes - 2 = No");
      System.in.read(inputValue);
      System.out.println("Has the string been reversed? 1 = Yes - 2 = No [ " + inputValue + " ]");

      if (inputValue == 1)
         {

         newInputString = "";
         for (i=intInputLength-1; i>=0; i--)
            {
            newInputString = newInputString + inputString[i];
            }

         inputString = newInputString;
         }

      System.out.println("Please wait decrypting...");

      newInputString = "";
      for (i=0; i<=intInputLength; i++)
         {
         for (n=0; n<=intArrayLength; n++)
            {
            if (inputString[i] == array[n])
               {
               x = (n-5);
               if (x <= 0)
                  {
                  x = x + intArrayLength;
                  newInputString = newInputString + array[x];
                  }
               else
                  {
                  newInputString = newInputString + array[x];
                  }
               }
            }
         }

      outputString = newInputString;
      System.out.print("Decrypted String: [ " + outputString + " ]");
      System.out.println();

      inputString = "";
      newInputString = "";



      }
   }

我已经编辑了上面的代码。在改变“Azodious”建议之后我现在得到的结果。

当我加密“cat999”时，我得到“hfyeeef”的结果“f”来自何处？当我解密“hfyeeef”时，我得到“ct995”的结果，“a”发生了什么，“5”是加密中附加“f”的结果。

Answer 1

如果n + 5 is >= array.length，则只需减去array.length：

int indexOfLetterToAppend = n + 5;
if (indexOfLetterToAppend >= array.length) {
    indexOfLetterToAppend -= array.length;
}

同样，如果n - 5 < 0，则只需添加array.length：

int indexOfLetterToAppend = n - 5;
if (indexOfLetterToAppend < 0) {
    indexOfLetterToAppend += array.length;
}

请注意，在循环中连接到String效率非常低。它会创建临时String和StringBuilder实例的大量内容。使用StringBuilder追加到最后，然后在toString()上致电StringBuilder。

另请注意，如果array[index]为array，则无法使用String。您必须使用array.charAt(index)，或将String转换为char数组。

Answer 2

让我们开始你的第一个for循环：

array = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ 0123456789";   

newInputString = ""; 

for (i=0; i<=intInputLength; i++)          
{          \
    for (n=0; n<=intArrayLength; n++)             
    {             
        if (inputString[i] == array[n])                
        {                
          newInputString = newInputString + array[(n+5)];                
        }
    } 
}

如果你给inputString = "56789";在n == 32 array[(n+5)]变为array[37]。（在条件范围内）

索引37肯定比array.length更多，你得到空字符串。

所以，当你发现(n+5) > array.length计算如下：

int diff =（n + 5） - array.length
newInputString = newInputString + array [diff-1]

在n-5获得价值时应执行的类似检查。

字符串的Java模算术

2 个答案: