生成多个列表中的所有组合

Question

鉴于未知数量的列表，每个列表的长度未知，我需要生成一个包含所有可能唯一组合的单个列表。 例如，给出以下列表：

X: [A, B, C] 
Y: [W, X, Y, Z]

然后我应该能够生成12种组合：

[AW, AX, AY, AZ, BW, BX, BY, BZ, CW, CX, CY, CZ]

如果添加了3个元素的第三个列表，我将有36个组合，依此类推。

关于如何用Java做到这一点的任何想法？
（伪代码也没问题）

Answer 1

你需要递归：

假设您的所有列表都在Lists，这是一个列表列表。让结果成为所需排列的列表：这样做

void GeneratePermutations(List<List<Character>> Lists, List<String> result, int depth, String current)
{
    if(depth == Lists.size())
    {
       result.add(current);
       return;
     }

    for(int i = 0; i < Lists.get(depth).size(); ++i)
    {
        GeneratePermutations(Lists, result, depth + 1, current + Lists.get(depth).get(i));
    }
}

最终的电话将是这样的

GeneratePermutations(Lists, Result, 0, EmptyString);

Answer 2

这个话题派上了用场。我用Java完全重写了以前的解决方案，更加用户友好。此外，我使用集合和泛型来提高灵活性：

/**
 * Combines several collections of elements and create permutations of all of them, taking one element from each
 * collection, and keeping the same order in resultant lists as the one in original list of collections.
 * 
 * <ul>Example
 * <li>Input  = { {a,b,c} , {1,2,3,4} }</li>
 * <li>Output = { {a,1} , {a,2} , {a,3} , {a,4} , {b,1} , {b,2} , {b,3} , {b,4} , {c,1} , {c,2} , {c,3} , {c,4} }</li>
 * </ul>
 * 
 * @param collections Original list of collections which elements have to be combined.
 * @return Resultant collection of lists with all permutations of original list.
 */
public static <T> Collection<List<T>> permutations(List<Collection<T>> collections) {
  if (collections == null || collections.isEmpty()) {
    return Collections.emptyList();
  } else {
    Collection<List<T>> res = Lists.newLinkedList();
    permutationsImpl(collections, res, 0, new LinkedList<T>());
    return res;
  }
}

/** Recursive implementation for {@link #permutations(List, Collection)} */
private static <T> void permutationsImpl(List<Collection<T>> ori, Collection<List<T>> res, int d, List<T> current) {
  // if depth equals number of original collections, final reached, add and return
  if (d == ori.size()) {
    res.add(current);
    return;
  }

  // iterate from current collection and copy 'current' element N times, one for each element
  Collection<T> currentCollection = ori.get(d);
  for (T element : currentCollection) {
    List<T> copy = Lists.newLinkedList(current);
    copy.add(element);
    permutationsImpl(ori, res, d + 1, copy);
  }
}

我正在使用guava库来创建集合。

Answer 3

此操作称为cartesian product。 Guava为其提供了实用功能：Lists.cartesianProduct

Answer 4

没有递归唯一组合：

    String sArray[] = new String []{"A", "A", "B", "C"};
    //convert array to list
    List<String> list1 = Arrays.asList(sArray);
    List<String> list2 = Arrays.asList(sArray);
    List<String> list3 = Arrays.asList(sArray);

    LinkedList<List <String>> lists = new LinkedList<List <String>>();

    lists.add(list1);
    lists.add(list2);
    lists.add(list3);

    Set<String> combinations = new TreeSet<String>();
    Set<String> newCombinations;

    for (String s: lists.removeFirst())
        combinations.add(s);

    while (!lists.isEmpty()) {
        List<String> next = lists.removeFirst();
        newCombinations =  new TreeSet<String>();
        for (String s1: combinations) 
            for (String s2 : next) 
              newCombinations.add(s1 + s2);               

        combinations = newCombinations;
    }
    for (String s: combinations)
        System.out.print(s+" ");    

Answer 5

为列表的通用列表List<List<T>>添加基于迭代器的答案，从Ruslan Ostafiichuk的答案中扩展了这个想法。我遵循的想法是：

     * List 1: [1 2]
     * List 2: [4 5]
     * List 3: [6 7]
     * 
     * Take each element from list 1 and put each element 
     * in a separate list.
     * combinations -> [ [1] [2] ]
     * 
     * Set up something called newCombinations that will contains a list
     * of list of integers
     * Consider [1], then [2]
     * 
     * Now, take the next list [4 5] and iterate over integers
     * [1]
     *  add 4   -> [1 4]
     *      add to newCombinations -> [ [1 4] ]
     *  add 5   -> [1 5]
     *      add to newCombinations -> [ [1 4] [1 5] ]
     * 
     * [2]
     *  add 4   -> [2 4]
     *      add to newCombinations -> [ [1 4] [1 5] [2 4] ]
     *  add 5   -> [2 5]
     *      add to newCombinations -> [ [1 4] [1 5] [2 4] [2 5] ]
     * 
     * point combinations to newCombinations
     * combinations now looks like -> [ [1 4] [1 5] [2 4] [2 5] ]
     * Now, take the next list [6 7] and iterate over integers
     *  ....
     *  6 will go into each of the lists
     *      [ [1 4 6] [1 5 6] [2 4 6] [2 5 6] ]
     *  7 will go into each of the lists
     *      [ [1 4 6] [1 5 6] [2 4 6] [2 5 6] [1 4 7] [1 5 7] [2 4 7] [2 5 7]]

现在的代码。我使用Set只是为了摆脱任何重复。可以用List替换。一切都应该无缝地工作。 ：）

public static <T> Set<List<T>> getCombinations(List<List<T>> lists) {
    Set<List<T>> combinations = new HashSet<List<T>>();
    Set<List<T>> newCombinations;

    int index = 0;

    // extract each of the integers in the first list
    // and add each to ints as a new list
    for(T i: lists.get(0)) {
        List<T> newList = new ArrayList<T>();
        newList.add(i);
        combinations.add(newList);
    }
    index++;
    while(index < lists.size()) {
        List<T> nextList = lists.get(index);
        newCombinations = new HashSet<List<T>>();
        for(List<T> first: combinations) {
            for(T second: nextList) {
                List<T> newList = new ArrayList<T>();
                newList.addAll(first);
                newList.add(second);
                newCombinations.add(newList);
            }
        }
        combinations = newCombinations;

        index++;
    }

    return combinations;
}

一个小试块..

public static void main(String[] args) {
    List<Integer> l1 = Arrays.asList(1,2,3);
    List<Integer> l2 = Arrays.asList(4,5);
    List<Integer> l3 = Arrays.asList(6,7);

    List<List<Integer>> lists = new ArrayList<List<Integer>>();
    lists.add(l1);
    lists.add(l2);
    lists.add(l3);

    Set<List<Integer>> combs = getCombinations(lists);
    for(List<Integer> list : combs) {
        System.out.println(list.toString());
    }

}

Answer 6

使用此处其他答案提供的嵌套循环解决方案来组合两个列表。

如果您有两个以上的列表，

1. 将前两个合并为一个新列表。
2. 将结果列表与下一个输入列表合并。
3. 重复。

Answer 7

• 没有递归
• 已订购
• 可以通过索引获得特定组合（不构建所有其他排列）：

最后的类和public class TwoDimensionalCounter<T> { private final List<List<T>> elements; public TwoDimensionalCounter(List<List<T>> elements) { this.elements = Collections.unmodifiableList(elements); } public List<T> get(int index) { List<T> result = new ArrayList<>(); for(int i = elements.size() - 1; i >= 0; i--) { List<T> counter = elements.get(i); int counterSize = counter.size(); result.add(counter.get(index % counterSize)); index /= counterSize; } return result;//Collections.reverse() if you need the original order } public int size() { int result = 1; for(List<T> next: elements) result *= next.size(); return result; } public static void main(String[] args) { TwoDimensionalCounter<Integer> counter = new TwoDimensionalCounter<>( Arrays.asList( Arrays.asList(1, 2, 3), Arrays.asList(1, 2, 3), Arrays.asList(1, 2, 3) )); for(int i = 0; i < counter.size(); i++) System.out.println(counter.get(i)); } } 方法：

'KVG'

Answer 8

使用 Java 8 Stream map 和 reduce 方法生成组合：

public static <T> List<List<T>> combinations(List<List<T>> lists) {
    // incorrect incoming data
    if (lists == null) return Collections.emptyList();
    return lists.stream()
            // non-null and non-empty lists
            .filter(list -> list != null && list.size() > 0)
            // represent each list element as a singleton list
            .map(list -> list.stream().map(Collections::singletonList)
                    // Stream<List<List<T>>>
                    .collect(Collectors.toList()))
            // summation of pairs of inner lists
            .reduce((list1, list2) -> list1.stream()
                    // combinations of inner lists
                    .flatMap(inner1 -> list2.stream()
                            // merge two inner lists into one
                            .map(inner2 -> Stream.of(inner1, inner2)
                                    .flatMap(List::stream)
                                    .collect(Collectors.toList())))
                    // list of combinations
                    .collect(Collectors.toList()))
            // otherwise an empty list
            .orElse(Collections.emptyList());
}

public static void main(String[] args) {
    List<String> list1 = Arrays.asList("A", "B", "C");
    List<String> list2 = Arrays.asList("W", "X", "Y", "Z");
    List<String> list3 = Arrays.asList("L", "M", "K");

    List<List<String>> lists = Arrays.asList(list1, list2, list3);
    List<List<String>> combinations = combinations(lists);

    // column-wise output
    int rows = 6;
    IntStream.range(0, rows).forEach(i -> System.out.println(
            IntStream.range(0, combinations.size())
                    .filter(j -> j % rows == i)
                    .mapToObj(j -> combinations.get(j).toString())
                    .collect(Collectors.joining(" "))));
}

按列输出：

[A, W, L] [A, Y, L] [B, W, L] [B, Y, L] [C, W, L] [C, Y, L]
[A, W, M] [A, Y, M] [B, W, M] [B, Y, M] [C, W, M] [C, Y, M]
[A, W, K] [A, Y, K] [B, W, K] [B, Y, K] [C, W, K] [C, Y, K]
[A, X, L] [A, Z, L] [B, X, L] [B, Z, L] [C, X, L] [C, Z, L]
[A, X, M] [A, Z, M] [B, X, M] [B, Z, M] [C, X, M] [C, Z, M]
[A, X, K] [A, Z, K] [B, X, K] [B, Z, K] [C, X, K] [C, Z, K]

---

^{另见：Cartesian product of an arbitrary number of sets}

Answer 9

您需要实施的操作称为笛卡尔积。 有关详细信息，请参阅https://en.wikipedia.org/wiki/Cartesian_product

我建议使用我的开源库，它可以完全满足您的需求： https://github.com/SurpSG/Kombi

有一个例子如何使用它： https://github.com/SurpSG/Kombi#usage-for-lists-1

注意： 该库专为高性能目的而设计。您可以观察banchmarks结果here

该库为您提供了非常好的吞吐量和持续的内存使用

Answer 10

以下是使用位掩码的示例。没有递归和多个列表

static List<Integer> allComboMatch(List<Integer> numbers, int target) {
    int sz = (int)Math.pow(2, numbers.size());
    for (int i = 1; i < sz; i++) {
        int sum = 0;
        ArrayList<Integer> result = new ArrayList<Integer>();
        for (int j = 0; j < numbers.size(); j++) {
            int x = (i >> j) & 1;
            if (x == 1) {
                sum += numbers.get(j);
                result.add(j);
            }
        }
        if (sum == target) {
            return result;
        }
    }
    return null;
}