Question

考虑一组包含N个数字的数字S，我想要S的2个子集（S1，S2），使得S1 U S2 = S. S1交叉点S2 = null 。我想要所有可能的S1和S2组合。我知道，如果我的Set包含N个数字，我将获得2 ^ N个子集组合。我试过以下代码。

DemoPartition Class

public class DemoPartition {

    ArrayList nodes = new ArrayList();
    ArrayList<Partitioning> partition = new ArrayList<>();

    public static void main(String[] args) {
        DemoPartition dp = new DemoPartition();
        try {
            int nums=3;

            dp.DividePartition(dp.nodes,nums);
            System.out.println(dp.partition);
            System.out.println(dp.partition.size());

        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    private void DividePartition(ArrayList nodes,int nums) {
         for (int i = 1; i <= nums; i++) {
                nodes.add(i);
            }
        for (int i = 0; i < nodes.size(); i++) {
            Partitioning parts = new Partitioning(new ArrayList(), new ArrayList());
            parts.getMobile().add(nodes.get(i));
            partition.add(parts);
        }

        ArrayList<Partitioning> temparray = new ArrayList<>();
        for (Partitioning parts : partition) {
            for (Object obj : nodes) {
                if (!parts.getMobile().contains(obj)) {
                    parts.getCloud().add(obj);
                }
            }
            if (nodes.size() != 2) {
                Partitioning newone = new Partitioning(parts.getCloud(), parts.getMobile());
                temparray.add(newone);
            }
        }
        partition.addAll(temparray);

        Partitioning parts = new Partitioning(new ArrayList(), new ArrayList());
        for (int i = 0; i < nodes.size(); i++) {
            parts.getMobile().add(nodes.get(i));
        }
        partition.add(parts);
        Partitioning newone = new Partitioning(partition.get(partition.size() - 1).getCloud(), partition.get(partition.size() - 1).getMobile());
        partition.add(newone);
    }

}

分区类

public class Partitioning {

    private ArrayList mobile;
    private ArrayList cloud;

    public Partitioning(ArrayList mobile, ArrayList cloud) {
        this.mobile = mobile;
        this.cloud = cloud;
    }

    public ArrayList getCloud() {
        return cloud;
    }

    public void setCloud(ArrayList cloud) {
        this.cloud = cloud;
    }

    public ArrayList getMobile() {
        return mobile;
    }

    public void setMobile(ArrayList mobile) {
        this.mobile = mobile;
    }

    @Override
    public String toString() {
        return "S1=" + mobile + " S2=" + cloud + "\n";
    }



}

nums = 3时输出

[S1=[1] S2=[2, 3]
, S1=[2] S2=[1, 3]
, S1=[3] S2=[1, 2]
, S1=[2, 3] S2=[1]
, S1=[1, 3] S2=[2]
, S1=[1, 2] S2=[3]
, S1=[1, 2, 3] S2=[]
, S1=[] S2=[1, 2, 3]
]
8

当nums为2和3时有效，当nums＆gt; 3时无效。我知道我的代码不完整，我没有得到通用逻辑如何解决这个问题，请帮帮我。

使用某些条件将一个集合划分为两个分区

0 个答案: