Question

请注意以下代码：

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

template <typename T>
void print_2d_vector(std::vector<std::vector<T>>& v)
{
  for(int i = 0; i < v.size(); i++)
  {
    std::cout << "{";
    for(int j = 0; j < v[i].size(); j++)
    {
      std::cout << v[i][j];
      if(j != v[i].size() - 1)
      {
        std::cout << ", ";
      }
    }
    std::cout << "}\n";
  }
}

template <typename T>
struct permcomb2
{
  std::vector<std::vector<T>> end_set;
  std::vector<T>* data;
  permcomb2(std::vector<T>& param) : data(&param) {}

  void helpfunc(std::vector<T>& seen, int depth)
  {
    if(depth == 0)
    {
      end_set.push_back(seen);
    }
    else
    {
      for(int i = 0; i < (*data).size(); i++)
      {
        seen.push_back((*data)[i]);
        helpfunc(seen, depth - 1);
        seen.pop_back();
      }
    }
  }
};

template <typename T>
std::vector<std::vector<T>> permtest(std::vector<T>& data, int subset_size)
{
  permcomb2<T> helpstruct(data);
  std::vector<T> empty {};
  helpstruct.helpfunc(empty, subset_size);
  return helpstruct.end_set;
}

using namespace std;
int main()
{
  std::vector<std::string> flavors {"Vanilla", "Chocolate", "Strawberry"};
  auto a1 = permtest(flavors, 2);

  cout << "Return all combinations with repetition\n";
  print_2d_vector(a1);
  return 0;
}

运行此代码会产生以下输出：

Return all combinations with repetition
{Vanilla, Vanilla}
{Vanilla, Chocolate}
{Vanilla, Strawberry}
{Chocolate, Vanilla}
{Chocolate, Chocolate}
{Chocolate, Strawberry}
{Strawberry, Vanilla}
{Strawberry, Chocolate}
{Strawberry, Strawberry}

注意这段代码不会做它声称做的事情！它不是返回所有重复给定子集大小（目标）的组合，而是返回所有排列，重复给定的子集大小。当然，获得组合的一种方法是如我所做的那样生成所有排列，然后循环以除去除了彼此排列之外的所有排列之外的所有排列。但我相信这绝对不是最有效的方法。

我已经看到了使用嵌套for循环来实现这一目标的方法，但是那些假设子集大小是提前知道的。我试图将它概括为任何子集大小，因此我试图以递归方式进行。问题是我不太确定如何更改递归的“helpfunc”以便以有效的方式生成所有组合。

为了澄清，预期的输出将是：

Return all combinations with repetition
{Vanilla, Vanilla}
{Vanilla, Chocolate}
{Vanilla, Strawberry}
{Chocolate, Chocolate}
{Chocolate, Strawberry}
{Strawberry, Strawberry}

那么如何更改我的代码以获得重复的所有组合而不是以有效的方式进行排列？

Answer 1

确保helpfunc循环从我们所在的索引开始，并仅考虑前面的索引。我们不想要的，因为它们只会是重复的。

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

template <typename T>
void print_2d_vector(std::vector<std::vector<T>>& v)
{
    for(int i = 0; i < v.size(); i++)
    {
        std::cout << "{";
        for(int j = 0; j < v[i].size(); j++)
        {
            std::cout << v[i][j];
            if(j != v[i].size() - 1)
            {
                sizetd::cout << ", ";
            }
        }
        std::cout << "}\n";
    }
}

template <typename T>
struct permcomb2
{
    std::vector<std::vector<T>> end_set;
    std::vector<T>& data;
    permcomb2(std::vector<T>& param) : data(param) {}

    void helpfunc(std::vector<T>& seen, int depth, int current) // Add one more param for the starting depth of our recursive calls
    {
        if(depth == 0)
        {
            end_set.push_back(seen);
        }
        else
        {
            for(int i = current; i < data.size(); i++) // Set the loop to start at given value
            {
                seen.push_back(data[i]);
                helpfunc(seen, depth - 1, i);
                seen.pop_back();
            }
        }
    }
};

template <typename T>
std::vector<std::vector<T>> permtest(std::vector<T>& data, int subset_size)
{
    permcomb2<T> helpstruct(data);
    std::vector<T> empty {};
    helpstruct.helpfunc(empty, subset_size, 0); // Initialize the function at depth 0
    return helpstruct.end_set;
}

using namespace std;
int main()
{
    std::vector<std::string> flavors {"Vanilla", "Chocolate", "Strawberry"};
    auto a1 = permtest(flavors, 2);

    cout << "Return all combinations with repetition\n";
    print_2d_vector(a1);
    return 0;
}

Answer 2

你可以考虑通过嵌套for循环解决这个问题，其中每个循环的计数器从前一个索引变为data大小。

for (int i = 0; i < data.size(); i++) {
  for (int j = i; j < data.size(); j++) {
    for (int k = j; k < data.size(); k++) {
      // etc...
  }
}

问题是循环嵌套的深度等于subset_size。我们可以通过在循环中进行递归调用来模拟这种任意深度的嵌套：

template <class T>
void solution(std::vector<T>& data, std::vector<std::vector<T>>& sol, int subset_size, int start=0, int depth=0) {
  if (depth == subset_size) return;

  // Assume that the last element of sol is a base vector
  // on which to append the data elements after "start"
  std::vector<T> base = sol.back();

  // create (data.size() - start) number of vectors, each of which is the base vector (above)
  // plus each element of the data after the specified starting index
  for (int i = start; i < data.size(); ++i) {
    sol.back().push_back(data[i]);                   // Append i'th data element to base 
    solution(data, sol, subset_size, i, depth + 1);  // Recurse, extending the new base
    if (i < data.size() - 1) sol.push_back(base);    // Append another base for the next iteration
  }
}

template <typename T>
std::vector<std::vector<T>> permtest(std::vector<T>& data, int subset_size) {
  std::vector<std::vector<T>> solution_set;
  solution_set.push_back(std::vector<T>());
  solution(data, solution_set, subset_size);
  return solution_set;
}

递归生成给定子集大小（C ++）

2 个答案: