从方法返回迭代器

Question

我似乎无法在Iterator从方法返回时打印元素removeTwos()。我试图从列表中删除只有两个字符的元素。

public class Main {

    public static void main(String[] args) {

        // write your code here

        List<String> list = new ArrayList<>();

        list.add("hi");

        list.add("what");

        list.add("who");

        list.add("ok");

        System.out.println(removeTwos(list));

    }

    public static String removeTwos(List<String> stringList) {

        Iterator<String> itr = stringList.iterator();

        for(int i = 0; i < stringList.size(); i++) {

            if(itr.hasNext() && itr.next().length() == 2) {

                itr.remove();

                System.out.println(itr.toString());

            }

        }

        return itr.toString();

    }

}

Answer 1

问题是：你根本不需要Iterator你想要做的事情。您可以使用列表的方法一一搜索列表中的每个项目。试试这个代码，看看它是否适合你：public class JavaApplication255 {&nbsp; &nbsp; public static void main(String[] args) {&nbsp; &nbsp; &nbsp; &nbsp; // write your code here&nbsp; &nbsp; &nbsp; &nbsp; List<String> list = new ArrayList<>();&nbsp; &nbsp; &nbsp; &nbsp; list.add("hi");&nbsp; &nbsp; &nbsp; &nbsp; list.add("what");&nbsp; &nbsp; &nbsp; &nbsp; list.add("who");&nbsp; &nbsp; &nbsp; &nbsp; list.add("ok");&nbsp; &nbsp; &nbsp; &nbsp; removeTwos(list);&nbsp; &nbsp; &nbsp; &nbsp; System.out.println(list);&nbsp; &nbsp; }&nbsp; &nbsp; public static void removeTwos(List<String> stringList){&nbsp; &nbsp; &nbsp; &nbsp; for(int i = stringList.size() - 1; i >= 0 ; i--){&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; String string = stringList.get(i);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (string.length() == 2){&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; stringList.remove(string);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp;}

Answer 2

要Iterator针对此问题使用 ，您可以将removeTwos方法编辑为如下所示：public static String removeTwos(List<String> stringList) {&nbsp; &nbsp; Iterator<String> itr = stringList.iterator();&nbsp; &nbsp; while (itr.hasNext()) {&nbsp; &nbsp; &nbsp; &nbsp; String value = itr.next();&nbsp; &nbsp; &nbsp; &nbsp; if (value.length() == 2) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.println(value);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; itr.remove();&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; return stringList.toString();}使用 时Iterator，在循环遍历列表时从列表中删除元素是安全的。这是一个证明它是安全的链接。

Answer 3

除非您需要使用 iter，否则使用流和过滤器很容易 1 班轮。List<String>&nbsp;noTwos&nbsp;=&nbsp;list.stream().filter(s->&nbsp;s.length()&nbsp;!=&nbsp;2).collect(Collectors.toList());