我正在嘗試找到與給定謂詞匹配的列表的第一個（任何）成員，如下所示：Item item = items.parallelStream()  .map(i -> i.doSomethingExpensive())  .filter(predicate)  .findAny()  .orElse(null);我希望一旦findAny()獲得匹配，它會立即返回，但情況似乎并非如此。相反，它似乎在返回之前等待 map 方法在大多數(shù)元素上完成。如何立即返回第一個結果并取消其他并行流？有沒有比使用諸如 的流更好的方法來做到這一點CompletableFuture？這是一個顯示行為的簡單示例：private static void log(String msg) {    private static void log(String msg) {    SimpleDateFormat sdf = new SimpleDateFormat("HH:mm:ss.SSS");    System.out.println(sdf.format(new Date()) + " " + msg);}Random random = new Random();List<Integer> nums = Arrays.asList(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14);Optional<Integer> num = nums.parallelStream()  .map(n -> {    long delay = Math.abs(random.nextLong()) % 10000;    log("Waiting on " + n + " for " + delay + " ms");    try { Thread.sleep(delay); }    catch (InterruptedException e) { System.err.println("Interruption error"); }    return n * n;  })  .filter(n -> n < 30)  .peek(n -> log("Found match: " + n))  .findAny();log("First match: " + num);日志輸出：14:52:27.061 Waiting on 9 for 2271 ms14:52:27.061 Waiting on 2 for 1124 ms14:52:27.061 Waiting on 13 for 547 ms14:52:27.061 Waiting on 4 for 517 ms14:52:27.061 Waiting on 1 for 1210 ms14:52:27.061 Waiting on 6 for 2646 ms14:52:27.061 Waiting on 0 for 4393 ms14:52:27.061 Waiting on 12 for 5520 ms14:52:27.581 Found match: 1614:52:27.582 Waiting on 3 for 5365 ms14:52:28.188 Found match: 414:52:28.275 Found match: 114:52:31.457 Found match: 014:52:32.950 Found match: 914:52:32.951 First match: Optional[0]一旦找到匹配項（在本例中為 16），findAny()不會立即返回，而是阻塞直到其余線程完成。在這種情況下，調(diào)用者在找到匹配后返回之前要額外等待 5 秒。