我能否以某种方式使用 Java8 流分析上一个和/或下一个元素?
例如,我可以计算相同的相邻数字吗?
public class Merge {
public static void main(String[] args) {
Stream<Integer> stream = Stream.of(0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1);
// How to get 3, 2, 2, 4 from above
}
}
最佳答案
如果你想让它变得懒惰,你必须通过 Stream.iterator()
或 Stream.spliterator()
来转义 Stream API。
否则,方法是使用自定义收集器调用终端操作 Stream.collect(Collector)
,这将消耗整个流。
@Test
public void test() {
Stream<Integer> input = Stream.of(0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1);
UniqCountSpliterator uniqCountSpliterator = new UniqCountSpliterator(input.spliterator());
long[] output = uniqCountSpliterator.stream()
.toArray();
long[] expected = {3, 2, 2, 4};
assertArrayEquals(expected, output);
}
import java.util.Spliterator;
import java.util.function.LongConsumer;
import java.util.stream.Stream;
import java.util.stream.StreamSupport;
public class UniqCountSpliterator implements Spliterator.OfLong {
private Spliterator wrapped;
private long count;
private Object previous;
private Object current;
public UniqCountSpliterator(Spliterator wrapped) {
this.wrapped = wrapped;
}
public LongStream stream() {
return StreamSupport.longStream(this, false);
}
@Override
public OfLong trySplit() {
return null;
}
@Override
public long estimateSize() {
return Long.MAX_VALUE;
}
@Override
public int characteristics() {
return NONNULL | IMMUTABLE;
}
@Override
public boolean tryAdvance(LongConsumer action) {
while (wrapped.tryAdvance(next -> current = next) && (null == previous || current.equals(previous))) {
count++;
previous = current;
}
if (previous == null) {
return false;
}
action.accept(count);
count = 1;
previous = null;
return true;
}
}
关于java - Java8流的 "merge"元素有可能吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39229767/