Java - ArrayList 元素的排列(整数) - 无法使其正常工作 [英] Java - Permutation of ArrayList elements (Integer) - Can't get it to work properly

问题描述

我一直在四处寻找以解决我的问题.我解决了很多问题，但这个问题仍然困扰着我 :S 我已经很长时间没有接触 Java 编程(一般编程)了，所以请理解那里！;)

I've been looking around quite a bit to solve my issue. I got many problems solved but this one is still haunting me :S It's been a long time I haven't touch Java programming (programming in general) so be understanding out there! ;)

我的目标是从整数数组中获得所有可能的组合.当我使用以下代码应用于整数 {1, 2, 3, 4} 的测试数组时，我希望有:
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
2 1 3 4
2 1 4 3
(...)
但这是我得到的
1 2 3 4
1 2 3 4 4 3
1 2 3 4 4 3 3 2 4

My goal is to get all the combination possible out of an array of integers. When I use the following code, applied to the test array of integer {1, 2, 3, 4}, I expect to have:
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
2 1 3 4
2 1 4 3
(...)
but here is what I get
1 2 3 4
1 2 3 4 4 3
1 2 3 4 4 3 3 2 4

有没有人有线索、建议甚至解决方案?提前致谢！

Does anybody have a clue, a suggestion or even a solution? Thanks in advance!

public class Calculation{
(...)
  public void Permute(ArrayList<Integer> soFar,ArrayList<Integer> rest){
    if(rest.isEmpty())    this.fillMatrice(convertIntegers(soFar)); // there it goes in a previously created arrow of int
    else{
        for(int k=0;k<rest.size();k++){
            ArrayList<Integer> next=new ArrayList<Integer>();
            next=soFar;
            next.add(rest.get(k));
            ArrayList<Integer> remaining=new ArrayList<Integer>();
            List<Integer> sublist = rest.subList(0, k);
            for(int a=0;a<sublist.size();a++)   remaining.add(sublist.get(a));
            sublist = rest.subList(k+1,rest.size());
            for(int a=0;a<sublist.size();a++)   remaining.add(sublist.get(a));
            Permute(next,remaining);
        }
    }
}
public static ArrayList<Integer> convertArray(int[] integers){
    ArrayList<Integer> convArray=new ArrayList<Integer>();
    for(int i=0;i<integers.length;i++)  convArray.add(integers[i]);
    return convArray;
}
public static int[] convertIntegers(List<Integer> integers){
    int[] ret = new int[integers.size()];
    for(int i=0;i<ret.length;i++)   ret[i]=integers.get(i).intValue();
    return ret;
}
public Calculation() {
    (...)
    ArrayList<Integer> soFar=new ArrayList<Integer>();
    int[] test={1,2,3,4};
    Permute(soFar,convertArray(test));
}

推荐答案

你可以试试Recursion来解决这个问题:

You can try Recursion to solve this issue:

public static void printPermutations(int[] n, int[] Nr, int idx) {
    if (idx == n.length) {  //stop condition for the recursion [base clause]
        System.out.println(Arrays.toString(n));
        return;
    }
    for (int i = 0; i <= Nr[idx]; i++) { 
        n[idx] = i;
        printPermutations(n, Nr, idx+1); //recursive invokation, for next elements
    }
}

可以从此链接获得更多信息:组合学:生成所有状态"——数组组合

More info can be had from this link: Combinatorics: generate all "states" - array combinations

您也可以在这里复制相同的逻辑.

You can replicate the same logic here as well.

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