简体   繁体   中英

Java PowerSet -prints only one

Having a problem, only printing one instead of each powerset and I can't quite spot it. Trying to write a program that gives out a powerset of a given set.

import java.util.HashSet;
import java.util.Iterator;


public class PowerSet {

    public static void main(String[] args) {

        HashSet<String> set = new HashSet<String>();
        HashSet<HashSet<String>> powerset;

        set.add("a");
        set.add("b");
        set.add("c");
        set.add("d");

        powerset = powerset(set);

        System.out.println(powerset);
    }

    public static HashSet<HashSet<String>> powerset( HashSet<String> set){

        if (set.size()== 0) {
            HashSet<HashSet<String>> pset = new HashSet<HashSet<String>>();
            HashSet<String> emptySet = new HashSet<String>();
            pset.add(emptySet);
            return pset;
        } else {
            HashSet<String> tmp;
            Iterator<String> it = set.iterator();
            String elt = it.next();
            set.remove(elt);
            HashSet<HashSet<String>> oldpset = powerset(set);
            HashSet<HashSet<String>> newpset = new HashSet<HashSet<String>>();
            Iterator<HashSet<String>> psetit = oldpset.iterator();
            while(psetit.hasNext()){
                tmp = psetit.next();
                newpset.add(tmp);
                tmp.add(elt);
                newpset.add(tmp);
            }

            return newpset;
        }
    }
}

One issue I see is that you remove the element elt from set before calling powerset(set) . Because Java objects act as pointers, when you remove an element from set at for the current call to powerset , you are modifying all references to it (even references stored on the stack from previous calls).

Another issue I see is that you use set.remove(elt) , you should be using it.remove() or you will mess up the iterator.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM