是否可以使用Streams.intRange函数？

实际上，JDK中再也没有这种方法了。您能获得的下一个最接近的位置是，

IntStream.range()

一种解决方案是实施您自己的解决方案

Spliterator.OfInt

public final class StepRange    implements Spliterator.OfInt{    private final int start;    private final int end;    private final int step;    private int currentValue;    public StepRange(final int start, final int end, final int step)    {        this.start = start;        this.end = end;        this.step = step;        currentValue = start;    }    @Override    public OfInt trySplit()    {        return null;    }    @Override    public long estimateSize()    {        return Long.MAX_VALUE;    }    @Override    public int characteristics()    {        return Spliterator.IMMUTABLE | Spliterator.DISTINCT;    }    @Override    public boolean tryAdvance(final IntConsumer action)    {        final int nextValue = currentValue + step;        if (nextValue > end) return false;        action.accept(currentValue);        currentValue = nextValue;        return true;    }}

然后，您将使用

StreamSupport.intStream()