从地图内的列表中删除重复的元素

Question

假设我有以下地图："A": [1, 2, 3, 4]"B": [5, 6, 1, 7]"C": [8, 1, 5, 9]如何从数组中删除重复的元素，以便返回仅包含从未重复的元素的映射？"A": [2, 3, 4]"B": [6, 7]"C": [8, 9]

Answer 1

您可能想这样做：

// Initializing the map

Map<String, List<Integer>> map = new LinkedHashMap<String, List<Integer>>() {

    {

        put("A", new ArrayList<>(Arrays.asList(1, 2, 3, 4)));

        put("B", new ArrayList<>(Arrays.asList(5, 6, 1, 7)));

        put("C", new ArrayList<>(Arrays.asList(8, 1, 5, 9)));

    }

};

// finding the common elements

List<Integer> allElements = map.values().stream().flatMap(List::stream).collect(Collectors.toList());

Set<Integer> allDistinctElements = new HashSet<>();

Set<Integer> commonElements = new HashSet<>();

allElements.forEach(element -> {

    if(!allDistinctElements.add(element)) {

        commonElements.add(element);

    }

});

// removing the common elements

map.forEach((key, list) -> list.removeAll(commonElements));

// printing the map

map.forEach((key, list) -> System.out.println(key + " = " + list));

输出：

A = [2, 3, 4]

B = [6, 7]

C = [8, 9]

Answer 2

首先你必须计算每个列表中的数字

Map<Integer, Long> countMap = map.values().stream()
            .flatMap(List::stream)
            .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

然后过滤其中 count == 1

Map<String, List<Integer>> result = map.entrySet().stream()
            .collect(Collectors.toMap(Map.Entry::getKey, e -> e.getValue().stream()
                    .filter(i -> countMap.get(i) == 1).collect(Collectors.toList())));