假設(shè)我有以下地圖："A": [1, 2, 3, 4]"B": [5, 6, 1, 7]"C": [8, 1, 5, 9]如何從數(shù)組中刪除重復(fù)的元素，以便返回僅包含從未重復(fù)的元素的映射？"A": [2, 3, 4]"B": [6, 7]"C": [8, 9]
2 回答
白衣染霜花
TA貢獻(xiàn)1796條經(jīng)驗(yàn) 獲得超10個(gè)贊
您可能想這樣做：
// 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));
輸出：
A = [2, 3, 4]
B = [6, 7]
C = [8, 9]
繁花如伊
TA貢獻(xiàn)2012條經(jīng)驗(yàn) 獲得超12個(gè)贊
首先你必須計(jì)算每個(gè)列表中的數(shù)字
Map<Integer, Long> countMap = map.values().stream() .flatMap(List::stream) .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
然后過(guò)濾其中 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())));
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