N个列表的排列 [英] Permutation of N lists

问题描述

我需要N个列表中的所有置换，直到程序启动我才知道N，这是我的SSCCE(我已经实现了建议给我的算法，但是它有一些错误).

i need all permutation from N lists, i dont know N until the program start, here is my SSCCE (i have implemented algorithm which was adviced to me, but it has some bugs).

首先，创建Place类:

First , create Place class:

public class Place {
public List<Integer> tokens ;

//constructor
public Place() {

  this.tokens = new ArrayList<Integer>();  
}

}

然后测试课程:

public class TestyParmutace {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here

        List<Place> places = new ArrayList<Place>();

        Place place1 = new Place();
        place1.tokens.add(1);
        place1.tokens.add(2);
        place1.tokens.add(3);
        places.add(place1); //add place to the list

        Place place2 = new Place();
        place2.tokens.add(3);
        place2.tokens.add(4);
        place2.tokens.add(5);
        places.add(place2); //add place to the list

        Place place3 = new Place();
        place3.tokens.add(6);
        place3.tokens.add(7);
        place3.tokens.add(8);
        places.add(place3); //add place to the list


        //so we have
        //P1 = {1,2,3}
        //P2 = {3,4,5}
        //P3 = {6,7,8}


        List<Integer> tokens = new ArrayList<Integer>();

        Func(places,0,tokens);

    }

    /**
     * 
     * @param places list of places
     * @param index index of current place
     * @param tokens list of tokens
     * @return true if we passed guard, false if we did not
     */


    public static boolean Func( List<Place> places, int index, List<Integer> tokens) 

    {

        if (index >= places.size())
            {

                // if control reaches here, it means that we've recursed through a particular combination
                // ( consisting of exactly 1 token from each place ), and there are no more "places" left



                String outputTokens = "";
                for (int i = 0; i< tokens.size(); i++) {

                    outputTokens+= tokens.get(i) +",";
                }
                System.out.println("Tokens: "+outputTokens);

                if (tokens.get(0) == 4 && tokens.get(1) == 5 && tokens.get(2) == 10) {
                    System.out.println("we passed the guard with 3,5,8");
                    return true;
                }

                else {
                    tokens.remove(tokens.get(tokens.size()-1));
                    return false;
                }

            }

        Place p = places.get(index);

        for (int i = 0; i< p.tokens.size(); i++)
            {

                tokens.add(p.tokens.get(i));
                //System.out.println("Pridali sme token:" + p.tokens.get(i));

                if ( Func( places, index+1, tokens ) ) return true;

            }
        if (tokens.size()>0)
            tokens.remove(tokens.get(0));

        return false;

    }
}

这是此代码的输出:

Tokens: 1,3,6,
Tokens: 1,3,7,
Tokens: 1,3,8,
Tokens: 3,4,6,
Tokens: 3,4,7,
Tokens: 3,4,8,
Tokens: 4,5,6,
Tokens: 4,5,7,
Tokens: 4,5,8,
Tokens: 2,3,6,
Tokens: 2,3,7,
Tokens: 2,3,8,
Tokens: 3,4,6,
Tokens: 3,4,7,
Tokens: 3,4,8,
Tokens: 4,5,6,
Tokens: 4,5,7,
Tokens: 4,5,8,
Tokens: 3,3,6,
Tokens: 3,3,7,
Tokens: 3,3,8,
Tokens: 3,4,6,
Tokens: 3,4,7,
Tokens: 3,4,8,
Tokens: 4,5,6,
Tokens: 4,5,7,
Tokens: 4,5,8,

因此，您看到，有些组合是正确的(1,3,6)，有些组合是不正确的(4,5,8)，而有些组合则是完全缺失的(2,4,8，..)如何解决此问题?位置的数量以及令牌的数量可能会有所不同，我只使用了3个位置，因为有2个位置可以工作，但有更多的位置却是越野车. 谢谢.

So, you see, some combinations are correct (1,3,6), some are incorrect (4,5,8) and some are completely missing (2,4,8,..) how to solve this problem ? number of places and also number of tokens in places can vary, i just used 3 places since with 2 places its working, but with more places it is buggy. Thanks.

推荐答案

您的算法几乎是正确的.我认为您在获得true时不需要返回true或false并停止当前迭代.我修改了您的方法Func:

You algorithm is almost correct. I think you don't need to return true or false and stop current iteration when you get true. I modified your method Func:

public static void Func( List<Place> places, int index, Deque<Integer> tokens) {
    if (index == places.size()) {
        // if control reaches here, it means that we've recursed through a particular combination
        // ( consisting of exactly 1 token from each place ), and there are no more "places" left
        String outputTokens = "";
        for (int token : tokens) {
            outputTokens += token + ",";
        }
        System.out.println("Tokens: "+outputTokens);
    } else {
        Place p = places.get(index);
        for (int token : p.tokens) {
            tokens.addLast(token);
            Func(places, index+1, tokens);
            token.removeLast();
        }
    }
}

我使用了 Deque ，因为它提供了方便的removeLast方法来删除最后添加的令牌.您可以将LinkedList作为Deque的实现传递.

I used Deque because it offers handy removeLast method to remove last added token. You can pass LinkedList as implementation of Deque.

更新

List<List<Integer>> combinations;

// Instead of printing result:
List<Integer> copy = new ArrayList<Integer>(tokens);
combinations.add(copy);

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