如何从LinkedHashMap<String，ArrayList的值进行组合<ArrayList>>

Question

假设我们在下面有LinkedHashMap<字符串，数组列表<数组列表>gt；

    {FRA=[[1, 2], [3, 4]], MEL=[[5, 6]]}

输出应为

    [1,2,5,6], [3,4,5,6]

类似地，如果输入为：

    {SFO=[[1]], SEA=[[2], [3], [4]], PHX=[[5], [6]]}

则预期输出为

    [1,2,5],[1,2,6],[1,3,5],[1,3,6],[1,4,5],[1,4,6]

我尝试过下面的代码，但没有得到预期的结果。

    public static ArrayList<List<String>> getCombinations(ArrayList<ArrayList<String>> valueSetList) {
    int comboCount = 1;
    for (List<String> valueSet : valueSetList)
        comboCount = Math.multiplyExact(comboCount, valueSet.size()); // Fail if overflow
    ArrayList<List<String>> combinations = new ArrayList<>(comboCount);
    for (int comboNumber = 0; comboNumber < comboCount; comboNumber++) {
        List<String> combination = new ArrayList<>(valueSetList.size());
        int remain = comboNumber;
        for (List<String> valueSet : valueSetList) {
            combination.add(valueSet.get(remain % valueSet.size()));
            
            remain /= valueSet.size();
        }
        combinations.add(combination);
    }
    return combinations;
}

非常感谢您的帮助。

Answer 1

以下递归解决方案基于Philipp Meister对关于笛卡尔乘积构建的类似问题的回答：

static List<List<Integer>> merge(List<List<List<Integer>>> lists) {
    List<List<Integer>> resultLists = new ArrayList<>();
    if (lists.size() == 0) {
        resultLists.add(new ArrayList<>());
    } else {
        List<List<Integer>> firstList = lists.get(0);
        List<List<Integer>> remainingLists = merge(lists.subList(1, lists.size()));
        for (List<Integer> first : firstList) {
            for (List<Integer> remaining : remainingLists) {
                List<Integer> resultList = new ArrayList<>();
                resultList.addAll(first);
                resultList.addAll(remaining);
                resultLists.add(resultList);
            }
        }
    }
    return resultLists;
}  

主要区别在于返回类型和根据要求将first列表的所有元素添加到嵌套列表中。

测试设置

private static void testCombinations(Map<String, List<List<Integer>>> map) {
    System.out.println("input map: " + map);
    
    List<List<Integer>> list = merge(new ArrayList<>(map.values()));
    
    list.forEach(System.out::println);
}
// --------
Map<String, List<List<Integer>>> map1 = new LinkedHashMap<>();
map1.put("SFO", Arrays.asList(Arrays.asList(1)));
map1.put("SEA", Arrays.asList(Arrays.asList(2), Arrays.asList(3), Arrays.asList(4)));
map1.put("PHX", Arrays.asList(Arrays.asList(5), Arrays.asList(6)));
testCombinations(map1);
        
Map<String, List<List<Integer>>> map2 = new LinkedHashMap<>();
map2.put("FRA", Arrays.asList(Arrays.asList(1, 2), Arrays.asList(3, 4)));
map2.put("MEL", Arrays.asList(Arrays.asList(5, 6)));
testCombinations(map2);

输出：

input map: {SFO=[[1]], SEA=[[2], [3], [4]], PHX=[[5], [6]]}
[1, 2, 5]
[1, 2, 6]
[1, 3, 5]
[1, 3, 6]
[1, 4, 5]
[1, 4, 6]
input map: {FRA=[[1, 2], [3, 4]], MEL=[[5, 6]]}
[1, 2, 5, 6]
[3, 4, 5, 6]

---

该解决方案可以使用流和递归flatMap链来实现(基于Marco13的答案(：

static <T extends List> Stream<List<T>> ofCombinations(List<? extends Collection<T>> mapValues, List<T> current) {
    return mapValues.isEmpty() ? Stream.of(current) :
        mapValues.get(0).stream().flatMap(list -> {
            List<T> combination = new ArrayList<T>(current);
            combination.addAll(list);
            return ofCombinations(mapValues.subList(1, mapValues.size()), combination);
        });
}
private static void testStreamCombinations(Map<String, List<List<Integer>>> map) {
    System.out.println("input map: " + map);
    
    ofCombinations(new ArrayList<>(map.values()), Collections.emptyList())
        .forEach(System.out::println);   
}
// invoking test method
testStreamCombinations(map1);
testStreamCombinations(map2);

测试输出：同上。