[英]Permutation of N lists
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).我需要 N 个列表中的所有排列,直到程序开始我才知道 N,这是我的 SSCCE(我已经实现了建议给我的算法,但它有一些错误)。
First, create Place class:首先,创建 Place class:
public class Place {
public List<Integer> tokens ;
//constructor
public Place() {
this.tokens = new ArrayList<Integer>();
}
}
And then testing class:然后测试 class:
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;
}
}
and here is the output of this code:这是此代码的 output:
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,
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?所以,你看,有些组合是正确的(1,3,6),有些是不正确的(4,5,8),有些完全没有(2,4,8,..)如何解决这个问题? 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.
地方的数量和地方的令牌数量可能会有所不同,我只使用了 3 个地方,因为有 2 个地方可以正常工作,但是更多的地方是有问题的。 Thanks.
谢谢。
You algorithm is almost correct.你的算法几乎是正确的。 I think you don't need to return
true
or false
and stop current iteration when you get true
.我认为您不需要返回
true
或false
并在获得true
时停止当前迭代。 I modified your method Func
:我修改了你的方法
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();
}
}
}
I used Deque because it offers handy removeLast
method to remove last added token.我使用Deque是因为它提供了方便的
removeLast
方法来删除最后添加的标记。 You can pass LinkedList
as implementation of Deque
.您可以将
LinkedList
作为Deque
的实现传递。
Update更新
List<List<Integer>> combinations;
// Instead of printing result:
List<Integer> copy = new ArrayList<Integer>(tokens);
combinations.add(copy);
You are trying to do set cross products, not really permutations.您正在尝试设置交叉产品,而不是真正的排列。 So you need to do
所以你需要做
for(Integer token1 : place1.tokens){
for(Integer token2 : place2.tokens){
for(Integer token3 : place3.tokens){
//crossValue = (token1, token2, token3);
}
}
}
Now if you want to have all permutations, for example for [1,3,6,]
also have [1,6,3]
, [3,6,1]
, etcc.. you need a function which output permutations given a list or array, see Permutation of array现在,如果你想拥有所有排列,例如
[1,3,6,]
也有[1,6,3]
, [3,6,1]
等。你需要一个 function 其中 output 排列给定列表或数组,参见数组的排列
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.