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Question

方法是

private List<HighWay> getWillSave(List<HighWay> total,
        HighWayRepository repository) {

    List<HighWay> willSave = new ArrayList<HighWay>();
    int lastSaved = total.size() - 1;
    while (lastSaved >= 0
            && repository.exists(total.get(lastSaved).getId())) {
        lastSaved--;
    }
    willSave.addAll(total.subList(0, ++lastSaved));
    return willSave;
}

现在我试着像这样改成stream，但我不知道这种改是否正确：

private List<HighWay> getWillSave(Stream<HighWay> notDealItem,
        HighWayRepository repository) {

    List<HighWay> willSave = new ArrayList<HighWay>();
    long lastSaved = notDealItem.count()-1;
    while (lastSaved >= 0
            && repository.exists(notDealItem.skip(lastSaved-1).findFirst().get().getId())) {
        lastSaved--;
    }
    willSave.addAll(notDealItem.collect(Collectors.toList()).subList(0, (int)(++lastSaved)));
    return willSave;
}

如何避免出现例外，解决问题？

我应该以某种方式重构代码吗？

Answer 1

好吧，我想你可以做这样的事...但与我们使用真正的函数式语言所能做的相比,它是非常不优雅的：

private List<HighWay> getWillSave(List<HighWay> total, HighWayRepository repository) {
    List<HighWay> willSave = new ArrayList<HighWay>();
    int size = total.size();
    int index = 
        IntStream.range(0,size)
            .mapToObj(i -> new Pair<>(i,size - i - 1))
            .filter(p -> !repository.exists(p.getValue().getId()))
            .findFirst()
            .orElseGet(() -> new Pair<>(0,null)).getKey();
    return total.subList(0,index);
}