Ç - 生成的x字符的话所有的可能性 [英] C - Generating all possibilities of X character words

问题描述

编辑：我的意思是置换，没有组合。谢谢你。

I meant Permutations, not combinations. Thanks.

我意识到这是一个相当开放式的问题，我不是在寻找每说code，但确实有些线索从哪里开始。我想是一个程序，它可以生成字符的每个组合为一个给定的长度，即用户输入4，和程序将生成的ASCII字符每个可能的组合为4的长度

I realize this is a rather open ended question, and I'm not looking for code per say, but really some hints where to start. What I want to make is a program, that can generate every combination of characters for a given length, i.e the user inputs 4, and the program will generate every possible combination of ASCII characters for a length of 4.

不能确定，我将开始，也许使用哈希表？当然，将需要的循环，但我不能确定如何设计他们生产的组合。截至目前为止，它一直是，循环的情况下，直到1000的事情已经发生了的例子。

Not really sure where I would start, perhaps the use of a hash table? Of course loops will be needed but I'm unsure how to design them to produce combinations. Up until now, its always been a case of, loop until 1000 things have happened for example.

任何意见得多AP preciated！

Any advice is much appreciated !

干杯，

T。

推荐答案

有关排列组合，你可以使用这样一个递归解决方案（这大概可以优化和改进）：

For permutations, you can use a recursive solution like this (which can probably be optimised and improved):

unordered_set<string> permute_string(int n) {    
    static const char chars[] = {
        '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', 'z'
    };

    unordered_set<string> s;

    if (n == 0) {
        s.insert("");

        return s;
    }

    unordered_set<string> perms = permute_string(n - 1);

    for (auto c = std::begin(chars); c < std::end(chars); ++c)
        for (auto i = perms.begin(); i != perms.end(); ++i)
            for (int pos = 0; pos < n; ++pos)
                s.insert(string(*i).insert(pos, 1, *c));

    return s;
}

请注意，这个函数的输出（无论你如何实现它）是的 26 ^N 的，这是456976时的 N 的（在该函数输入）为4。

Note that the output of this function (no matter how you implement it) is 26ⁿ, which is 456,976 when n (the input for this function) is 4.

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