具有Java传递字符串错误的暴力算法 [英] Brute Force Algorithm w/Java Passing String Error
问题描述
我认为这可能是一个非常简单的解决方案，但我不太确定。
I think this may be a very simple solution but I am not too sure.
我正在尝试创建一个字符数组，通过递增a来对它们进行排序数组的特定索引，并在最后将它们放入一个字符串中。
I am trying to create an array of characters, sort through them by incrementing a particular index of the array and throwing them into a string at the end.
我已经完成了这项工作，（通过将结果打印到控制台进行验证。aaaaaaaato aaaaaaab等等。但是，我的新版代码通过ASCII值33 - 126排序。'！'到'〜'
I have accomplished this, (verified via printing the result out to the console. "aaaaaaaa" to "aaaaaaab" and so on. However, my new version of the code sorts through ASCII Values 33 - 126. '!' to '~'
现在，一件事我想做的是在main方法中调用该字符串，以便我们分配的加密/解密程序可以一遍又一遍地使用该字符串。
Now, the one thing I am trying to do is call that string in the main method so that the encryption/decryption program that we were assigned, can use this string over and over again.
我们最初分配的程序可以在这里找到：
http://www.avajava.com/tutorials/lessons/how-do-i-crycrypt-and-decrypt-files-using-des.html
The program we were originally assigned can be found here: http://www.avajava.com/tutorials/lessons/how-do-i-encrypt-and-decrypt-files-using-des.html
import java.lang.Class.*;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import javax.crypto.Cipher;
import javax.crypto.CipherInputStream;
import javax.crypto.CipherOutputStream;
import javax.crypto.SecretKey;
import javax.crypto.SecretKeyFactory;
import javax.crypto.spec.DESKeySpec;
public class Example1 {
public String keyGen() {
//create an array used for storing each character
char array[] = new char[8];
//for loop checks for each character between '!' and '~'
for (char c0 = '!'; c0 <= '~'; c0++) {
array[0] = c0;
for (char c1 = '!'; c1 <= '~'; c1++) {
array[1] = c1;
for (char c2 = '!'; c2 <= '~'; c2++) {
array[2] = c2;
for (char c3 = '!'; c3 <= '~'; c3++) {
array[3] = c3;
for (char c4 = '!'; c4 <= '~'; c4++) {
array[4] = c4;
for (char c5 = '!'; c5 <= '~'; c5++) {
array[5] = c5;
for (char c6 = '!'; c6 <= '~'; c6++) {
array[6] = c6;
for (char c7 = '!'; c7 <= '~'; c7++) {
array[7] = c7;
//create new string that stores the array
String pKey = new String(array);
//trying to return the new string
return pKey;
}
}
}
}
}
}
}
}
}
public static void main(String []args) {
try {
// I am getting an error here; I know it has something to do with static references
String key = new String(keyGen(pKey);
// needs to be at least 8 characters for DES
FileInputStream fis = new FileInputStream("original.txt");
FileOutputStream fos = new FileOutputStream("encrypted.txt");
encrypt(key, fis, fos);
FileInputStream fis2 = new FileInputStream("encrypted.txt");
FileOutputStream fos2 = new FileOutputStream("decrypted.txt");
decrypt(key, fis2, fos2);
} catch (Throwable e) {
e.printStackTrace();
}
}
public static void encrypt(String key, InputStream is, OutputStream os) throws Throwable {
encryptOrDecrypt(key, Cipher.ENCRYPT_MODE, is, os);
}
public static void decrypt(String key, InputStream is, OutputStream os) throws Throwable {
encryptOrDecrypt(key, Cipher.DECRYPT_MODE, is, os);
}
public static void encryptOrDecrypt(String key, int mode, InputStream is, OutputStream os) throws Throwable {
DESKeySpec dks = new DESKeySpec(key.getBytes());
SecretKeyFactory skf = SecretKeyFactory.getInstance("DES");
SecretKey desKey = skf.generateSecret(dks);
Cipher cipher = Cipher.getInstance("DES"); // DES/ECB/PKCS5Padding for SunJCE
if (mode == Cipher.ENCRYPT_MODE) {
cipher.init(Cipher.ENCRYPT_MODE, desKey);
CipherInputStream cis = new CipherInputStream(is, cipher);
doCopy(cis, os);
} else if (mode == Cipher.DECRYPT_MODE) {
cipher.init(Cipher.DECRYPT_MODE, desKey);
CipherOutputStream cos = new CipherOutputStream(os, cipher);
doCopy(is, cos);
}
}
public static void doCopy(InputStream is, OutputStream os) throws IOException {
byte[] bytes = new byte[64];
int numBytes;
while ((numBytes = is.read(bytes)) != -1) {
os.write(bytes, 0, numBytes);
}
os.flush();
os.close();
is.close();
}
}
Thi
推荐答案
希望这可以帮助您解决问题。至于你现在，你的 keyGen（）等同于以下代码：
Hope this helps you solve the problem. As for right now, your keyGen() is equivalent to the following code:
public String keyGen() {
return "!!!!!!!!";
}
因为您在进入最里面的<$ c $后立即返回值c> for 循环。也许您想要更改方法以返回 List< String> 并将最内层的中的字符串添加到循环中该列表，并在所有 for 循环之后返回该列表？这样你就可以遍历 main 方法中的列表。
because you return the value as soon as you enter the innermost for loop. Maybe you want to change the method to return a List<String> and add your strings in the innermost for loop to that list, and return that list after all of the for loops? This way you could iterate through the list in your main method.
如果我计算正确，则有93 ^ 8 = 5.595.818.096.650.401个不同的字符串。将所有这些存储在列表中是不好的建议。 Dukeling在评论中指出，最好使用自定义 Iterator< String> 。
If I have counted correct, there are 93^8 = 5.595.818.096.650.401 different strings. It was bad advice to store all those in a list. Dukeling noted in the comments, that it would be better to use a custom Iterator<String> for that.
编辑
以下是这种迭代器的实现：
Here's an implementation of such an iterator:
import java.util.Arrays;
import java.util.Iterator;
public class BruteForceIterator implements Iterator<String> {
private char min, max;
private char[] current;
private char[] last;
private int reachedLast = 0;
public BruteForceIterator(char min, char max, int length) {
this.min = min;
this.max = max;
current = new char[length];
Arrays.fill(current, min);
last = new char[length];
Arrays.fill(last, max);
}
@Override
public boolean hasNext() {
return reachedLast < 2;
}
@Override
public String next() {
String str = new String(current);
for(int i = current.length - 1; i >= 0; i--) {
char next = following(current[i]);
current[i] = next;
if (next != min) {
break;
}
}
if (Arrays.equals(current, last) || reachedLast > 0) {
reachedLast++;
}
return str;
}
private char following(char in) {
if (in < max) {
return (char) (in + 1);
} else {
return min;
}
}
@Override
public void remove() {
throw new UnsupportedOperationException("No with me, sir!");
}
public static void main(String[] args) {
BruteForceIterator bit = new BruteForceIterator('a', 'c', 3);
while (bit.hasNext()) {
System.out.println(bit.next());
}
}
}
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