// returns [i,j] where lists.get(i).get(j) equals o,or null if o is not present.
public static int[] indices(List<? extends List<?>> lists,Object o) {
    return IntStream.range(0,lists.size())
                    .Boxed()
                    .flatMap(i -> IntStream.range(0,lists.get(i).size()).mapToObj(j -> new int[]{i,j}))
                    .parallel()
                    .filter(a -> {
                        System.out.println(Arrays.toString(a));     // For testing only
                        return Objects.equals(o,lists.get(a[0]).get(a[1]));
                    })
                    .findAny()
                    .orElse(null);
}

当我运行以下代码时@H_301_6@

List<List<String>> lists = Arrays.asList(
        Arrays.asList("A","B","C"),Arrays.asList("D","E","F","G"),Arrays.asList("H","I"),Collections.nCopies(5,"J")
);
System.out.println("Indices are " + Arrays.toString(indices(lists,"J")));

输出是这样的@H_301_6@

[0,0]
[0,1]
[0,2]
[3,0]
[3,1]
[3,3]
[2,4]
[1,0]
[1,1]
[2,1]
[1,2]
[1,3]
Indices are [3,0]

换句话说,即使在找到对象之后搜索仍继续.是不是应该是一个短路操作？我错过了什么？此外,在迭代列表列表或锯齿状数组时,利用并行性的最佳方法是什么？@H_301_6@

编辑@H_301_6@

根据@ Sotirios的回答,我得到了一个输出@H_301_6@

Thread[ForkJoinPool.commonPool-worker-3,5,main] [3,0]
Thread[main,main] [2,1]
Thread[ForkJoinPool.commonPool-worker-1,main] [1,0]
Thread[ForkJoinPool.commonPool-worker-1,2]
Thread[ForkJoinPool.commonPool-worker-1,3]
Thread[main,main] [0,1]
Thread[ForkJoinPool.commonPool-worker-3,1]
Thread[main,2]
Thread[ForkJoinPool.commonPool-worker-3,3]
Thread[ForkJoinPool.commonPool-worker-3,4]
Indices are [3,0]

请注意@H_301_6@

Thread[ForkJoinPool.commonPool-worker-3,main]

在找到答案后继续搜索.@H_301_6@