Question

我有

List<String> listOne = new ArrayList<String>();
listOne.add("3");
listOne.add("11");
listOne.add("20");`

和

List<String> listTwo = new ArrayList<String>();
listTwo.add("1");
listTwo.add("2");
listTwo.add("3");
listTwo.add("6-11");
listTwo.add("18-20");

我想从listTwo中存在的listOne元素中删除。在此示例中，应删除的值包括：3，6-11和18-20。 6-11因为在listOne中我们的值11和18-20的值相同。

如果可能的话，你能帮助我吗？我不知道如何解决它。由于其删除规则，这与其他人不同。否则，另一个类似问题的其他答案并没有给我解决方案。

Answer 1

public class Test {
public static void main(String[] args) {
    List<String> listOne = new ArrayList<String>();
    listOne.add("3");
    listOne.add("11");
    listOne.add("20");

    List<String> listTwo = new ArrayList<String>();
    listTwo.add("1");
    listTwo.add("2");
    listTwo.add("3");
    listTwo.add("6-11");
    listTwo.add("18-20");

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

    for (String secondItem : listTwo) {
        boolean match = false;
        for (String firstItem : listOne) {
            if(secondItem.contains(firstItem)) {
                match = true;
            }
        }
        if(!match) {
            thirdList.add(secondItem);
        }
    }
    for (String string : thirdList) {
        System.out.println(string);
    }
}

}

Answer 2

注释表明只有明确列出的值才能导致元素被删除，例如：

listOne    listTwo             Action
"1"        "1", "3-9", "11"    remove "1", but not "11"
"2"        "1", "3-9", "11"    do nothing
"3"        "1", "3-9", "11"    remove "3-9"
"4"        "1", "3-9", "11"    do nothing (3-9 is not a range)

所以，有了这个奇怪的逻辑，这是一个例子：

public static void main(String[] args) {
    // should remove 3, 6-11, 18-20
    System.out.println(remove(Arrays.asList("3", "11", "20"),
                              new ArrayList<>(Arrays.asList("1", "2", "3", "6-11", "18-20"))));
    // should remove 2, not remove 6-11 (not a range), and not remove 18-20 (8 is not 18, 2 is not 20)
    System.out.println(remove(Arrays.asList("2", "8"),
                              new ArrayList<>(Arrays.asList("1", "2", "3", "6-11", "18-20"))));
}
private static List<String> remove(List<String> listOne, List<String> listTwo) {
    for (Iterator<String> listIter = listTwo.iterator(); listIter.hasNext(); ) {
        String value = listIter.next();
        if (shouldRemove(value, listOne))
            listIter.remove();
    }
    return listTwo; // for easy of use
}
private static boolean shouldRemove(String value, List<String> listOne) {
    int idx = value.indexOf('-');
    if (idx == -1) {
        for (String ref : listOne)
            if (ref.equals(value))
                return true;
    } else {
        String value1 = value.substring(0, idx);
        String value2 = value.substring(idx + 1);
        for (String ref : listOne)
            if (ref.equals(value1) || ref.equals(value2))
                return true;
    }
    return false;
}

输出

[1, 2]
[1, 3, 6-11, 18-20]

Answer 3

public static void main(String[] args) {

        List<String> listOne = new ArrayList<>();
        listOne.add("3");
        listOne.add("11");
        listOne.add("20");

        List<String> listTwo = new CopyOnWriteArrayList<>();
        listTwo.add("1");
        listTwo.add("2");
        listTwo.add("3");
        listTwo.add("6-11");
        listTwo.add("18-20");

        listOne.stream().forEach(value -> {
        listTwo.stream().filter(value2 -> (value2.contains(value))).forEach(value2 -> {
            String[] minMax = value2.split(value2);
            if (minMax.length == 1 && value2.equals(value)) {
                listTwo.remove(value2);
            } else if (minMax.length == 2) {
                int intValue2 = Integer.parseInt(value2);
                int min = Integer.parseInt(minMax[0]);
                int max = Integer.parseInt(minMax[1]);
                if (intValue2 >= min && intValue2 <= max) {
                    listTwo.remove(value2);
                }
            }
        });
        });

        listTwo.stream().forEach(System.out::println);
}

Lambda Filtering Collections

What is CopyOnWriteArrayList?

Answer 4

    List<String> listOne = new ArrayList<String>();
    listOne.add("3");
    listOne.add("11");
    listOne.add("20");
    List<String> listTwo = new ArrayList<String>();
    listTwo.add("1");
    listTwo.add("2");
    listTwo.add("3");
    listTwo.add("6-11");
    listTwo.add("18-20");

   List<String> values = new ArrayList<>();     // to store matched Strings

   for (String s : listTwo) {
        String strs[] = new String[2];
        if (s.contains("-")) {
            strs = s.split("-");                 // check for Strings
        }
        if (listOne.contains(s) 
               || listOne.contains(strs[0]) 
               || listOne.contains(strs[1])) {
            values.add(s);                       // save the value to a list
        }

    }
    listTwo.removeAll(values);

如何删除Java中另一个列表中存在的列表元素？

4 个答案: