嘗試在Java中可能的每個字母/文字組合

Question

我正在嘗試創建一個程序，它將反覆「創建」一系列字符，並將它們與關鍵字（用戶或計算機未知）進行比較。如果你願意的話，這與「蠻力」攻擊非常相似，除非這會邏輯地構建它所能包含的每一個字母。

另一件事是，我臨時構建了這段代碼來處理JUST 5個字母的單詞，並將它分解爲一個「值」二維字符串數組。我把它作爲一個非常臨時的解決方案，幫助我邏輯地發現我的代碼在做什麼，然後再將它放入超動態和複雜的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. 
} 
} 

我的目標是讓輸出是這樣的：（五年字母例如）

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

等等等等等等。現在通過運行這個代碼，這不是我所希望的。現在，它會去：

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

任何幫助表示讚賞。我真的很難解決這個難題。

Answer 1

首先，您的字母表丟失0（零）和z。它也有3兩次。

其次，使用36個可能字符的五個字母的數量是60,466,176。方程是(size of alphabet)^(length of word)。在這種情況下，即36^5。我運行了你的代碼，並且它只生成了176個排列組合。

在我的機器上，基本實現了五個嵌套for循環，每個遍歷字母表，它花費了144秒來生成和打印所有的排列。所以，如果你得到了快速的結果，你應該檢查生成的內容。

當然，手動嵌套循環並不是一個有效的解決方案，因爲當你希望單詞的長度是可變的，所以你仍然有一些工作要做。但是，我的建議是注意細節並驗證你的假設！

祝你好運。