从地图内的列表中删除重复的元素[关闭]

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，以便返回只包含从未重复的元素的地图？

"A": [2, 3, 4]
"B": [6, 7]
"C": [8, 9]

Answer 1

你可能想这样做：

// 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()));

然后在 count == 1 的地方过滤

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())));