Question

我正在尝试创建一个程序，它将一遍又一遍地“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，这非常类似于“蛮力”攻击，除了这将从逻辑上构建出每一个字母。

另一件事是，我暂时构建了这个代码来处理JUST 5字母单词，并将其分解为“值”2D字符串数组。我把它作为一个非常临时的解决方案，帮助从逻辑上发现我的代码正在做什么，然后再把它扔进超级动态和复杂的for循环中。

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.
}
}

我的目标是让输出看起来像这样:(对于五个字母的例子）

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

等等。但是现在通过运行此代码，它并不是我所希望的。现在，它将会：

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，您的字母表缺少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单词的数量是60,466,176。等式为(size of alphabet)^(length of word)。在这种情况下，即36^5。我运行了你的代码，它只产生176个排列。

在我的机器上，基本实现了五个嵌套for循环，每个循环遍历字母表，生成并打印所有排列需要144秒。因此，如果您获得快速结果，则应检查正在生成的内容。

当然，手动嵌套for循环并不是一个有效的解决方案，当你希望单词的长度可变时，你还有一些工作要做。但是，我的建议是关注细节并验证您的假设！

祝你好运。

尝试在Java中使用每个字母/单词组合

1 个答案: