在C ++中生成复合排列的绝佳方法

Question

I have a class Test which contains two vectors of a Letter class, a user defined type for which the less-than operator (<) has been implemented. How can I best generate all possible permutations of Test?

class Test
{
  vector<Letter> letter_box_a;
  vector<Letter> letter_box_b;
}

So if letter_box_a contains the letters A and B and letter_box_b contains C and D the valid permutations of Test would be (AB)(CD), (BA)(CD), (AB)(DC) and (BA)(DC).

Although I am able to brute force it, I was trying to write a better (more elegant/efficient) function which would internally call std::next_permutation on the underlying containers allowing me to do

Test test;
while (test.set_next_permutation())
{
    // Do the stuff
}

but it appears to be a bit trickier than I first anticipated. I don't necessarily need an STL solution but would like an elegant solution.

Answer 1

If you want to use std::next_permutation , you need a nested loop for each vector you are permuting:

std::string s0 = "ab";
std::string s1 = "cd";

do
{
    do
    {
        cout << s0 << "" << s1 << endl;
    } while (std::next_permutation(s0.begin(), s0.end()));
} while (std::next_permutation(s1.begin(), s1.end()));

Output:

abcd
bacd
abdc
badc

And, in the class:

class Foo
{
public:
    Foo(std::string_view arg_a, std::string_view arg_b)
        : a(arg_a)
        , b(arg_b)
        , last(false)
    { }

    void reset_permutations()
    {
        last = false;
    }

    bool next_permutation(std::string& r)
    {
        if (last)
            return false;

        if (not std::next_permutation(a.begin(), a.end()))
            if (not std::next_permutation(b.begin(), b.end()))
                last = true;

        r = a + b;
        return true;
    }

private:
    std::string a, b;
    bool last;
};

int main(int argc, const char *argv[])
{
    Foo foo("ab", "cd");
    string s;
    while (foo.next_permutation(s))
        cout << s << endl;
    return 0;
}

Answer 2

I would think you could do something like

bool Test::set_next_permutation() {
    auto &a = letter_box_a, &b = letter_box_b;  // entirely to shorten the next line
    return std::next_permutation(a.start(), a.end()) || std::next_permutation(b.start(), b.end());
}

(Of course, a while loop will skip the initial permutation in any case. You want a do ... while loop instead.)