n 组的所有可能组合

Question

我有n套。 每个 Set 具有不同数量的元素。 我想编写一个算法，为我提供集合中所有可能的组合。 例如，假设我们有：

S1={1,2}, S2={A,B,C}, S3={$,%,£,!}

组合应该看起来像

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

...等等。 可能的组合数为2*3*4 = 24

请帮我用Java编写这个算法。

Answer 1

递归是你的朋友：

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 + ",");
    }
  }
}

响应 OP 关于将其推广到对象集的问题：

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);
    }
  }
}

最后是一个迭代变体。 它基于增加一个计数器数组，当它达到集合的大小时，计数器会包装并携带：

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) + "}";
  }
}

Answer 2

如果有人想要矩阵而不是打印，我稍微修改了代码：

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;
  }
}

Answer 3

map 和 reduce方法

您可以创建一个通用方法来获取不同类型和数量的集合及其元素的笛卡尔积。

在线试试吧！

public static void main(String[] args) {
  Set<Integer> a = Set.of(1, 2);
  List<String> b = List.of("A", "B", "C");
  Set<Character> c = Set.of('$', '%', '£', '!');

  Set<Set<Object>> cpSet = cartesianProduct(HashSet::new, a, b, c);
  List<List<Object>> cpList = cartesianProduct(ArrayList::new, a, b, c);

  // output, order may vary
  System.out.println(cpSet);
  System.out.println(cpList);
}

/**
 * @param nCol the supplier of the output collection
 * @param cols the input array of collections
 * @param <R>  the type of the return collection
 * @return the cartesian product of the multiple collections
 */
@SuppressWarnings("unchecked")
public static <R extends Collection<?>> R cartesianProduct(
      Supplier nCol, Collection<?>... cols) {
  // check if supplier is not null
  if (nCol == null) return null;
  return (R) Arrays.stream(cols)
      // non-null and non-empty collections
      .filter(col -> col != null && col.size() > 0)
      // represent each element of a collection as a singleton collection
      .map(col -> (Collection<Collection<?>>) col.stream()
          .map(e -> Stream.of(e).collect(Collectors.toCollection(nCol)))
          .collect(Collectors.toCollection(nCol)))
      // summation of pairs of inner collections
      .reduce((col1, col2) -> (Collection<Collection<?>>) col1.stream()
          // combinations of inner collections
          .flatMap(inner1 -> col2.stream()
              // concatenate into a single collection
              .map(inner2 -> Stream.of(inner1, inner2)
                  .flatMap(Collection::stream)
                  .collect(Collectors.toCollection(nCol))))
          // collection of combinations
          .collect(Collectors.toCollection(nCol)))
      // otherwise an empty collection
      .orElse((Collection<Collection<?>>) nCol.get());
}

输出（裁剪后，顺序可能会有所不同）：

[[1, A, !], [1, !, B], [A, !, 2], [1, A, £], [1, !, C]...
[[1, A, $], [1, A, %], [1, A, £], [1, A, !], [1, B, $]...

---

^{另请参阅：在 Java 中查找笛卡尔积}

Answer 4

krilid 的出色解决方案，我对其进行了一些修改以返回所有组合的列表。

@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);
    }
}