尝试在Java中可能的每个字母/文字组合

Question

我正在尝试创建一个程序，它将反复“创建”一系列字符，并将它们与关键字（用户或计算机未知）进行比较。如果你愿意的话，这与“蛮力”攻击非常相似，除非这会逻辑地构建它所能包含的每一个字母。

另一件事是，我临时构建了这段代码来处理JUST 5个字母的单词，并将它分解为一个“值”二维字符串数组。我把它作为一个非常临时的解决方案，帮助我逻辑地发现我的代码在做什么，然后再将它放入超动态和复杂的for-loops。

public class Sample{ 

static String key, keyword = "hello"; 
static String[] list = {"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","1","2","3","3","4","5","6","7","8","9"}; 
int keylen = 5; // Eventually, this will be thrown into a for-loop, to get dynamic "keyword" sizes. (Will test to every word, more/less than 5 characters eventually) 

public static void main(String[] args) { 

    String[] values = {"a", "a", "a", "a", "a"}; // More temporary hardcodes. If I can figure out the for loop, the rest can be set to dynamic values. 

    int changeout_pos = 0; 
    int counter = 0; 

    while(true){ 
     if (counter == list.length){ counter = 0; changeout_pos++; } // Swap out each letter we have in list, in every position once. 

     // Try to swap them. (Try/catch is temporary lazy way of forcing the computer to say "we've gone through all possible combinations") 
     try { values[changeout_pos] = list[counter]; } catch (Exception e) { break; } 

     // Add up all the values in their respectful positions. Again, will be dynamic (and in a for-loop) once figured out. 
     key = values[0] + values[1] + values[2] + values[3] + values[4]; 
     System.out.println(key); // Temporarily print it. 
     if (key.equalsIgnoreCase(keyword)){ break; } // If it matches our lovely keyword, then we're done. We've done it! 

     counter ++; // Try another letter. 
    } 

    System.out.println("Done! \nThe keyword was: " + key); // Should return what "Keyword" is. 
} 
} 

我的目标是让输出是这样的：（五年字母例如）

aaaaa 
aaaab 
aaaac 
... 
aaaba 
aaabb 
aaabc 
aaabd 
... 
aabaa 
aabab 
aabac 
... 

等等等等等等。现在通过运行这个代码，这不是我所希望的。现在，它会去：

aaaaa 
baaaa 
caaaa 
daaaa 
... (through until 9) 
9aaaa 
9baaa 
9caaa 
9daaa 
... 
99aaa 
99baa 
99caa 
99daa 
... (Until it hits 99999 without finding the "keyword") 

任何帮助表示赞赏。我真的很难解决这个难题。

Answer 1

首先，您的字母表丢失0（零）和z。它也有3两次。

其次，使用36个可能字符的五个字母的数量是60,466,176。方程是(size of alphabet)^(length of word)。在这种情况下，即36^5。我运行了你的代码，并且它只生成了176个排列组合。

在我的机器上，基本实现了五个嵌套for循环，每个遍历字母表，它花费了144秒来生成和打印所有的排列。所以，如果你得到了快速的结果，你应该检查生成的内容。

当然，手动嵌套循环并不是一个有效的解决方案，因为当你希望单词的长度是可变的，所以你仍然有一些工作要做。但是，我的建议是注意细节并验证你的假设！

祝你好运。