blocks|key|1875446|text|如果流的大小未知，那么可能没有办法消耗整个流并缓冲到目前为止遇到的最后一个n元素。您可以使用某种类型的双端队列或专门的环形缓冲区自动保持其最大大小(有关某些实现，请参阅this+related+question+)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1875447|public+static+<T>+List<T>+lastN(Stream<T>+stream,+int+n)+{
++++Deque<T>+result+=+new+ArrayDeque<>(n);
++++stream.forEachOrdered(x+->+{
++++++++if+(result.size()+==+n)+{
++++++++++++result.pop();
++++++++}
++++++++result.add(x);
++++});
++++return+new+ArrayList<>(result);
}|code-block|syntax|javascript|1875448|所有这些操作(size，pop，add)的复杂度都应该为O(1)，因此长度为(未知)n的流的总体复杂度将为O(n)。|1875449|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/7266042/java-ring-buffer^0|11|1|2C|L|0|0|0|7|4|C|3|G|3|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

In case the stream has unknown size, there's probably no way around consuming the entire stream and buffering the last <code>n</code> elements encountered so far. You can do this using some kind of deque, or a specialized ring-buffer automatically maintaining its maximum size (see <a href="https://stackoverflow.com/questions/7266042/java-ring-buffer">this related question</a> for some implementations).

<pre><code>public static &lt;T&gt; List&lt;T&gt; lastN(Stream&lt;T&gt; stream, int n) {
 Deque&lt;T&gt; result = new ArrayDeque&lt;&gt;(n);
 stream.forEachOrdered(x -&gt; {
 if (result.size() == n) {
 result.pop();
 }
 result.add(x);
 });
 return new ArrayList&lt;&gt;(result);
}
</code></pre>

All of those operations (<code>size</code>, <code>pop</code>, <code>add</code>) should have complexity of O(1), so the overall complexity for a stream with (unknown) length n would be O(n).

blocks|key|1875456|text|自定义收集器可以这样编写：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1875457|public+static+<T>+Collector<T,+?,+List<T>>+lastN(int+n)+{
++++return+Collector.<T,+Deque<T>,+List<T>>of(ArrayDeque::new,+(acc,+t)+->+{
++++++++if(acc.size()+==+n)
++++++++++++acc.pollFirst();
++++++++acc.add(t);
++++},+(acc1,+acc2)+->+{
++++++++while(acc2.size()+<+n+&&+!acc1.isEmpty())+{
++++++++++++acc2.addFirst(acc1.pollLast());
++++++++}
++++++++return+acc2;
++++},+ArrayList::new);
}|code-block|syntax|javascript|1875458|并像这样使用它：|1875459|List<String>+lastTen+=+input.stream().collect(lastN(10));|1875460|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

A custom collector can be written like this:

<pre><code>public static &lt;T&gt; Collector&lt;T, ?, List&lt;T&gt;&gt; lastN(int n) {
 return Collector.&lt;T, Deque&lt;T&gt;, List&lt;T&gt;&gt;of(ArrayDeque::new, (acc, t) -&gt; {
 if(acc.size() == n)
 acc.pollFirst();
 acc.add(t);
 }, (acc1, acc2) -&gt; {
 while(acc2.size() &lt; n &amp;&amp; !acc1.isEmpty()) {
 acc2.addFirst(acc1.pollLast());
 }
 return acc2;
 }, ArrayList::new);
}
</code></pre>

And use it like this:

<pre><code>List&lt;String&gt; lastTen = input.stream().collect(lastN(10));
</code></pre>

blocks|key|1875477|text|有时我需要一个"oneliner“(在本例中是一个三行代码)，因为创建一个收集器太麻烦了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1875478|如果流很小，那么可以在不牺牲太多性能的情况下再次执行reverse、limit和reverse。这将产生最后n个元素。|offset|length|style|CODE|1875479|如果需要过滤，则它很有用，因为在这种情况下，无法指定大小。|1875480|Stream.of(1,+2,+3,+4,+5,+6,+7,+8,+9)
++.filter(i+->+i+%25+2+==+0)
++.sorted(Comparator.reverseOrder())
++.limit(2)
++.sorted(Comparator.naturalOrder())
++.forEach(System.out::println);+//+prints+6+8|code-block|syntax|javascript|1875481|entityMap^0|0|Q|7|Y|5|14|7|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]|$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|Z|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

Sometimes I need a "oneliner" (in this case a three liner) as creating a collector is just too much fuss.

If the stream is small then it is possible to <code>reverse</code>, <code>limit</code> and <code>reverse</code> again without much sacrificing performance. This will result the last n elements.

It is useful if filtering is required as in that case it is not possible to specify the size.

<pre><code>Stream.of(1, 2, 3, 4, 5, 6, 7, 8, 9)
 .filter(i -&gt; i % 2 == 0)
 .sorted(Comparator.reverseOrder())
 .limit(2)
 .sorted(Comparator.naturalOrder())
 .forEach(System.out::println); // prints 6 8
</code></pre>

I am wondering is there an alternative to

<pre><code>List&lt;X&gt; lastN = all.subList(Math.max(0, all.size() - n), all.size());
</code></pre>

with <a href="https://docs.oracle.com/javase/8/docs/api/java/util/stream/package-summary.html">stream</a> usage?

Get last n elements from stream

我想知道有没有别的办法List<X> lastN = all.subList(Math.max(0, all.size() - n), all.size());使用？

问从流中获取最后n个元素
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问从流中获取最后n个元素EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问从流中获取最后n个元素
EN