在 C++ 中获取积分器数组的所有可能组合

Question

I have an array of integers, for example: a[1,2,3] . I would like to get all possible combinations of these numbers where they don't repeat, possibly with recursion.

I saw something like this done with strings here: Get all combinations without duplicates but don't know how to adapt it to integers without using any standard algorithms.

So I would like something like this as an output: {1},{2},{3},{1,2},{2,3},{1,3},{1,2,3}

Thanks in advance!

Answer 1

You can achieve all permutations of a list with comparable elements using std::next_permutation from <algorithms> library.

The cppreference has a nice article about this: https://en.cppreference.com/w/cpp/algorithm/next_permutation

We can use the code example to apply to your question.

#include <algorithm>
#include <string>
#include <iostream>
#include <vector>

void print_vector(const std::vector<int>& array) {
    for (const int item : array)
    {
        std::cout << item << " ";
    }
    std::cout << std::endl;
}
 
int main()
{
    std::vector<int> a({ 5,1,8,7,2 });
    std::sort(a.begin(), a.end());
    do {
        print_vector(a);
    } while(std::next_permutation(a.begin(), a.end()));
}

Answer 2

You can use std::next_permuation to achieve your goal. Keep in mind you need to sort the array before starting to use this algorithm. The loop will exit the first time std::next_permuation returns false. If the array isn't sorted by the time you start std::next_permuation loop, you will miss all the arrays that are lexicographically lower than the current one when entering the loop.

    int main()
    {
        std::vector<int> a = { 5,1,8,7,2 };
        std::sort(a.begin(), a.end());
        std::cout << "Possible permutations :\n";
        do {
            for (auto o : a)
                std::cout << o;
            std::cout << std::endl;
        } while (std::next_permutation(a.begin(), a.end()));
        return 0;
    }