如何在不使用递归的情况下找到字符串的所有排列?

2022-01-24 00:00:00 递归 algorithm iteration c++

谁能帮我解决这个问题:这是一个查找任意长度字符串的所有排列的程序.需要相同的非递归形式.(首选C语言实现)

Can someone help me with this: This is a program to find all the permutations of a string of any length. Need a non-recursive form of the same. ( a C language implementation is preferred)

using namespace std;

string swtch(string topermute, int x, int y)
{
  string newstring = topermute;
  newstring[x] = newstring[y];
  newstring[y] = topermute[x]; //avoids temp variable
  return newstring;
}

void permute(string topermute, int place)
{
  if(place == topermute.length() - 1)
  {
    cout<<topermute<<endl;
  }
  for(int nextchar = place; nextchar < topermute.length(); nextchar++)
  {
    permute(swtch(topermute, place, nextchar),place+1);
  }
}

int main(int argc, char* argv[])
{    
  if(argc!=2)    
  {
    cout<<"Proper input is 'permute string'";
    return 1;
  }
  permute(argv[1], 0);
  return 0;    
}

推荐答案

基于堆栈的非递归等效代码:

A stack based non-recursive equivalent of your code:

#include <iostream>
#include <string>

struct State
{
    State (std::string topermute_, int place_, int nextchar_, State* next_ = 0)
        : topermute (topermute_)
        , place (place_)
        , nextchar (nextchar_)
        , next (next_)
    {
    }

    std::string topermute;
    int place;
    int nextchar;

    State* next;
};

std::string swtch (std::string topermute, int x, int y)
{
    std::string newstring = topermute;
    newstring[x] = newstring[y];
    newstring[y] = topermute[x]; //avoids temp variable

    return newstring;
}

void permute (std::string topermute, int place = 0)
{
    // Linked list stack.
    State* top = new State (topermute, place, place);

    while (top != 0)
    {
        State* pop = top;
        top = pop->next;

        if (pop->place == pop->topermute.length () - 1)
        {
            std::cout << pop->topermute << std::endl;
        }

        for (int i = pop->place; i < pop->topermute.length (); ++i)
        {
            top = new State (swtch (pop->topermute, pop->place, i), pop->place + 1, i, top);
        }

        delete pop;
    }
}

int main (int argc, char* argv[])
{
    if (argc!=2)    
    {
        std::cout<<"Proper input is 'permute string'";
        return 1;
    }
    else
    {
        permute (argv[1]);
    }

    return 0;
}

我试图让它像 C 一样，并避免使用 c++ STL 容器和成员函数(不过为了简单起见，使用了构造函数).

I've tried to make it C-like and avoided c++ STL containers and member functions (used a constructor for simplicity though).

请注意，排列是按与原始顺序相反的顺序生成的.

Note, the permutations are generated in reverse order to the original.

我应该补充一点，以这种方式使用堆栈只是模拟递归.

I should add that using a stack in this way is just simulating recursion.

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