假设我有以下地图：

"A": [1, 2, 3, 4]
"B": [5, 6, 1, 7]
"C": [8, 1, 5, 9]

如何清除数组以仅使用从不重复的元素返回地图。

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

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]