使用重复键合并数组的映射

我有两个数组映射.

Map> map1 = new HashMap<>();
Map> map2 = new HashMap<>();

我想将它们合并到一个新地图中.
如果两个映射中都存在密钥,那么我应该合并数组.

例如:

map1.put("k1", Arrays.asList("a0", "a1"));
map1.put("k2", Arrays.asList("b0", "b1"));

map2.put("k2", Arrays.asList("z1", "z2"));

// Expected output is 
Map 3: {k1=[a0, a1], k2=[b0, b1, z1, z2]}

我尝试用流来做到这一点

Map> map3 = Stream.of(map1, map2)
    .flatMap(map -> map.entrySet().stream())
    .collect(Collectors.toMap(
        Map.Entry::getKey,
        e -> e.getValue().stream().collect(Collectors.toList())
    ));

如果地图中没有相同的键,则此工作.否则,我得到例外

Exception in thread "main" java.lang.IllegalStateException: Duplicate key k2 (attempted merging values [b0, b1] and [z1, z2])
    at java.base/java.util.stream.Collectors.duplicateKeyException(Collectors.java:133)
    at java.base/java.util.stream.Collectors.lambda$uniqKeysMapAccumulator$1(Collectors.java:180)
    at java.base/java.util.stream.ReduceOps$3ReducingSink.accept(ReduceOps.java:169)
    at java.base/java.util.HashMap$EntrySpliterator.forEachRemaining(HashMap.java:1751)
    at java.base/java.util.stream.ReferencePipeline$Head.forEach(ReferencePipeline.java:658)
    at java.base/java.util.stream.ReferencePipeline$7$1.accept(ReferencePipeline.java:274)
    at java.base/java.util.Spliterators$ArraySpliterator.forEachRemaining(Spliterators.java:948)
    at java.base/java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:484)
    at java.base/java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:474)
    at java.base/java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:913)
    at java.base/java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234)
    at java.base/java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:578)
    at im.djm.Test.main(Test.java:25)

有没有办法用流完成这项任务？
或者我必须迭代地图？

在重复键的情况下使用合并功能:

Map> map3 = Stream.of(map1, map2)
                .flatMap(map -> map.entrySet().stream())
                .collect(Collectors.toMap(
                        Map.Entry::getKey,
                        e -> new ArrayList<>(e.getValue()),
                        (left, right) -> {left.addAll(right); return left;}
                ));

请注意,我已经改变 e -> e.getValue().stream().collect(Collectors.toList()),以new ArrayList<>(e.getValue())保证我们始终有一个可变的列表,我们可以在合并功能添加进去.

也许。但是，您更有可能通过使用迭代手动组合条目来使所有事情正确。我不知道是否还有其他人需要处理此代码，但他们可能会对这种易于阅读的方法表示感谢。