public static void main(String[] args) {long t0 = System.nanoTime();

        // 初始化一个范围100万整数流,求能被2整除的数字，toArray（）是终点方法

        int a[] = IntStream.range(0, 1_000_000)           .filter(p -> p % 2 == 0).toArray();

        long t1 = System.nanoTime();

        // 和上面功能一样，这里是用并行流来计算

        int b[] = IntStream.range(0, 1_000_000).parallel().filter(p -> p % 2 == 0).toArray();

        long t2 = System.nanoTime();

        // 我本机的结果是serial: 0.06s, parallel 0.02s，证明并行流确实比顺序流快

        System.out.printf("serial: %.2fs, parallel %.2fs%n", (t1 - t0) * 1e-9, (t2 - t1) * 1e-9);

} // serial: 0.07s, parallel 0.02s

List<Integer> numbers = new ArrayList<>(Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9));

numbers.parallelStream().forEachOrdered(out::print);