[英]Java modulus arithmetic for string
我正在为我的大学项目创建一个简单的Java程序,但无法创建数组字符串模数。 该程序只获取一个字符串,并针对数组中的字母从+5或-5对其进行加密或解密。 但是,如果我在输入字符串中输入数组(5,6,7,8,9)的最后5个中的任何一个,它将不返回任何内容,但应在开始时返回(a,b,c,d,e)字母的数组。
我必须使用的程序非常原始,只允许代码采用某种方式,因此为什么要这样编码。
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。 更改“ Azodious”建议后,我现在得到的结果。
当我加密“ cat999”时,我得到“ hfyeeef”的结果,“ f”从何而来? 当我解密“ hfyeeef”时,我得到“ ct995”的结果,“ a”发生了什么,“ 5”是加密中附加的“ f”的结果。
如果n + 5 is >= array.length
,则仅减去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,在循环中串联到String
效率很低。 它会创建临时String
和StringBuilder
实例的副本。 使用StringBuilder
追加到末尾,然后在StringBuilder
上调用toString()
。
另请注意,如果array
是String
,则不能使用array[index]
。 您必须使用array.charAt(index)
,或将String
转换为char数组。
让我们开始第一个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)];
}
}
}
如果您给inputString = "56789"
; 在n == 32
array[(n+5)]
变成array[37]
。 (在条件内)
索引37绝对比array.length
,您会得到空字符串。
因此,当您找到(n+5) > array.length
计算以下内容:
在n-5
取小牛肉时应执行的类似检查。
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