all possible combination of n sets

问题:

I have n Sets. each Set has different number of elements. I would like to write an algorithm which give me all possible combinations from the sets. for example: Lets say we have:

S1={1,2}, S2={A,B,C}, S3={$,%,£,!}

a combination should look like

C1={1,A,$}
C2={1,A,%}

.... and so on the number of possible combination will be 2*3*4 = 24

Please Help me to write this algorithm in Java.

Many thanks in Advance

回答1:

Recursion is your friend:

public class PrintSetComb{
    public static void main(String[] args){
        String[] set1 = {"1","2"};
        String[] set2 = {"A","B","C"};
        String[] set3 = {"$", "%", "£", "!"};
        String[][] sets = {set1, set2, set3};

        printCombinations(sets, 0, "");
    }

    private static void printCombinations(String[][] sets, int n, String prefix){
        if(n >= sets.length){
            System.out.println("{"+prefix.substring(0,prefix.length()-1)+"}");
            return;
        }
        for(String s : sets[n]){
            printCombinations(sets, n+1, prefix+s+",");
        }
    }
}

In response to question from OP about generalizing it to sets of Objects:

import java.util.Arrays;

public class PrintSetComb{

    public static void main(String[] args){
        Integer[] set1  = {1,2};
        Float[] set2    = {2.0F,1.3F,2.8F};
        String[] set3   = {"$", "%", "£", "!"};
        Object[][] sets = {set1, set2, set3};

        printCombinations(sets, 0, new Object[0]);
    }

    private static void printCombinations(Object[][] sets, int n, Object[] prefix){
        if(n >= sets.length){
            String outp = "{";
            for(Object o: prefix){
                outp = outp+o.toString()+",";
            }
            System.out.println(outp.substring(0,outp.length()-1)+"}");
            return;
        }
        for(Object o : sets[n]){
            Object[] newPrefix = Arrays.copyOfRange(prefix,0,prefix.length+1);
            newPrefix[newPrefix.length-1] = o;
            printCombinations(sets, n+1, newPrefix);
        }
    }
}

And finally an iterative variant. It is based on increasing an array of counters where the counter wraps and carries when it reaches the size of the set:

import java.util.Arrays;

public class PrintSetCombIterative{

    public static void main(String[] args){
            String[] set1 = {"1","2"};
            String[] set2 = {"A","B","C"};
            String[] set3 = {"$", "%", "£", "!"};
            Object[][] sets = {set1, set2, set3};

            printCombinations(sets);
    }


    private static void printCombinations(Object[][] sets){
        int[] counters = new int[sets.length];

        do{
           System.out.println(getCombinationString(counters, sets));
        }while(increment(counters, sets));
    }

    private static boolean increment(int[] counters, Object[][] sets){
            for(int i=counters.length-1;i>=0;i--){
                if(counters[i] < sets[i].length-1){
                    counters[i]++;
                    return true;
                } else {
                    counters[i] = 0;
                }
            }
            return false;
    }

    private static String getCombinationString(int[] counters, Object[][] sets){
        String combo = "{";
        for(int i = 0; i<counters.length;i++){
            combo = combo+sets[i][counters[i]]+",";
        }
        return combo.substring(0,combo.length()-1)+"}";

    }
}

回答2:

In the case someone wants the matrix instead of printing, I slightly modified the code:

public class TestSetCombinations2 {

    public static void main(String[] args) {
        Double[] set1 = {2.0,3.0};
        Double[] set2 = {4.0,2.0,1.0};
        Double[] set3 = {3.0, 2.0, 1.0, 5.0};
        Double[] set4 = {1.0,1.0};
        Object[][] sets = {set1, set2, set3, set4};

        Object[][] combinations = getCombinations(sets);

        for (int i = 0; i < combinations.length; i++) {
            for (int j = 0; j < combinations[0].length; j++) {
              System.out.print(combinations[i][j]+" ");
            }
            System.out.println();
        }
    }

    private static Object[][] getCombinations(Object[][] sets) {

      int[] counters = new int[sets.length];
        int count = 1;   
        int count2 = 0;

        for (int i = 0; i < sets.length; i++) {
          count *= sets[i].length;
        }

        Object[][] combinations = new Object[count][sets.length];

        do{
           combinations[count2++] = getCombinationString(counters, sets);
        } while(increment(counters, sets));

        return combinations;
    }

    private static Object[] getCombinationString(int[] counters, Object[][] sets) {

      Object[] o = new Object[counters.length];
      for(int i = 0; i<counters.length;i++) {
        o[i] = sets[i][counters[i]];
      }
      return o;

    }

    private static boolean increment(int[] counters, Object[][] sets) {
        for(int i=counters.length-1;i>=0;i--) {
            if(counters[i] < sets[i].length-1) {
                counters[i]++;
                return true;
            } else {
                counters[i] = 0;
            }
        }
        return false;
    }
}

回答3:

Great solution by krilid, I modified it a bit to return a list of all combinations.

@Test
public void testMap() throws Exception {
    char[] one = new char[]{'a', 'b', 'c'};
    char[] two = new char[]{'d', 'e', 'f'};
    char[] three = new char[]{'g', 'h', 'i', 'j'};
    char[][] sets = new char[][]{one, two, three};
    List<List<Character>> collector = new ArrayList<>();
    ArrayList<Character> combo = new ArrayList<>();
    combinations(collector, sets, 0, combo);
    System.out.println(collector);
}

private void combinations(List<List<Character>> collector, char[][] sets, int n, ArrayList<Character> combo) {
    if (n == sets.length) {
        collector.add(new ArrayList<>(combo));
        return;
    }
    for (char c : sets[n]) {
        combo.add(c);
        combinations(collector, sets, n + 1, combo);
        combo.remove(combo.size() - 1);
    }
}