与类推
IntSummaryStatistics，创建一个
PointStatistics收集所需信息的类。它定义了两种方法：一种用于记录a的值
Point，一种用于组合两个的值
Statistics。
class PointStatistics { private int minX = Integer.MAX_VALUE; private int maxX = Integer.MIN_VALUE; private int minY = Integer.MAX_VALUE; private int maxY = Integer.MIN_VALUE; public void accept(Point p) { minX = Math.min(minX, p.x); maxX = Math.max(maxX, p.x); minY = Math.min(minY, p.y); maxY = Math.max(minY, p.y); } public void combine(PointStatistics o) { minX = Math.min(minX, o.minX); maxX = Math.max(maxX, o.maxX); minY = Math.min(minY, o.minY); maxY = Math.max(maxY, o.maxY); } // getters}然后，您可以将收集
Stream<Point>到中
PointStatistics。
class Program { public static void main(String[] args) { List<Point> points = new ArrayList<>(); // populate 'points' PointStatistics statistics = points .stream() .collect(PointStatistics::new, PointStatistics::accept, PointStatistics::combine); }}更新
我完全对 OP得出的结论感到困惑，所以我决定编写JMH基准测试。
基准设置：
# JMH version: 1.21# VM version: JDK 1.8.0_171, Java HotSpot(TM) 64-Bit Server VM, 25.171-b11# Warmup: 1 iterations, 10 s each# Measurement: 10 iterations, 10 s each# Timeout: 10 min per iteration# Benchmark mode: Average time, time/op
对于每次迭代，我都生成大小为100K，1M，10M 的随机
Points（
new Point(random.nextInt(),random.nextInt())）共享列表。
结果是
100K
Benchmark Mode Cnt Score Error UnitscustomCollector avgt 10 6.760 ± 0.789 ms/opforEach avgt 10 0.255 ± 0.033 ms/opfourStreamsavgt 10 5.115 ± 1.149 ms/opstatistics avgt 10 0.887 ± 0.114 ms/optwoStreams avgt 10 2.869 ± 0.567 ms/op
1M
Benchmark Mode Cnt Score Error UnitscustomCollector avgt 10 68.117 ± 4.822 ms/opforEach avgt 10 3.939 ± 0.559 ms/opfourStreamsavgt 10 57.800 ± 4.817 ms/opstatistics avgt 10 9.904 ± 1.048 ms/optwoStreams avgt 10 32.303 ± 2.498 ms/op
10M
Benchmark Mode Cnt Score Error UnitscustomCollector avgt 10 714.016 ± 151.558 ms/opforEach avgt 10 54.334 ± 9.820 ms/opfourStreamsavgt 10 699.599 ± 138.332 ms/opstatistics avgt 10 148.649 ± 26.248 ms/optwoStreams avgt 10 429.050 ± 72.879 ms/op
