Java 中的抽象顺序列表子列表()方法，示例

原文:https://www . geesforgeks . org/abstractsequentialist-sublist-method-in-Java-with-example/

Java 中抽象顺序列表的子列表()方法用于获取此列表中指定的 fromIndex(包含)和 toIndex(不包含)之间的部分的视图。(如果 fromIndex 和 toIndex 相等，则返回的列表为空。)返回的列表由该列表支持，因此返回列表中的非结构性变化反映在该列表中，反之亦然。返回的列表支持该列表支持的所有可选列表操作。

语法:

protected List<E> subList(int fromIndex, 
                                int toIndex)

参数:这些方法取两个参数:

  • fromIndex: 要从中提取元素的起始索引。
  • 到索引:要从中提取元素的结束索引。(独家)

返回值:此方法返回此列表内指定范围的视图 异常:此方法抛出:

  • indexout of BoundSexception:如果端点索引值超出范围。
  • IllegalArgumentException :如果端点索引出现故障。

以下示例说明了 abstractsequentialilist . sublist()方法:

例 1 :

// Java program to demonstrate the
// working of subList() method

import java.util.*;

public class GFG {
    public static void main(String[] args)
    {

        // creating an AbstractSequentialList
        AbstractSequentialList<Integer> arr
            = new LinkedList<Integer>();

        // use add() method
        // to add values in the list
        arr.add(1);
        arr.add(2);
        arr.add(3);
        arr.add(12);
        arr.add(9);
        arr.add(13);

        // prints the list before removing
        System.out.println("AbstractSequentialList: "
                           + arr);

        // Getting subList of 1st 2 elements
        // using subList() method
        System.out.println("subList of 1st 2 elements: "
                           + arr.subList(0, 2));
    }
}

Output:

AbstractSequentialList: [1, 2, 3, 12, 9, 13]
subList of 1st 2 elements: [1, 2]

例 2:

// Java program to demonstrate the
// working of subList() method

import java.util.*;

public class GFG {
    public static void main(String[] args)
    {

        // creating an AbstractSequentialList
        AbstractSequentialList<Integer> arr
            = new LinkedList<Integer>();

        // use add() method
        // to add values in the list
        arr.add(1);
        arr.add(2);
        arr.add(3);
        arr.add(12);
        arr.add(9);
        arr.add(13);

        // prints the list before removing
        System.out.println("AbstractSequentialList: "
                           + arr);

        System.out.println("Trying to get "
                           + "subList of 11th elements: ");

        try {

            // Getting subList of 10th
            // using subList() method
            arr.subList(10, 11);
        }
        catch (Exception e) {
            System.out.println(e);
        }
    }
}

Output:

AbstractSequentialList: [1, 2, 3, 12, 9, 13]
Trying to get subList of 11th elements: 
java.lang.IndexOutOfBoundsException: toIndex = 11

版权属于：月萌API www.moonapi.com，转载请注明出处

本文链接：https://moonapi.com/news/18096.html