Java AbstractList subList()用法及代码示例

java.util.AbstractList 类的 subList() 方法用于返回此列表在指定 fromIndex(包括)和 toIndex(不包括)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更改会反映在该列表中，反之亦然。返回的列表支持所有可选的列表操作。

用法:

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

参数：此方法将以下参数作为参数。

fromIndex:subList 的低端点(包括)
toIndex:subList 的高端(不包括)

返回值：此方法返回此列表中指定范围的视图。

异常：此方法引发以下异常。

IndexOutOfBoundsException:如果端点索引值超出范围(fromIndex 大小)
IllegalArgumentException:如果端点索引乱序(fromIndex > toIndex)

以下是说明 subList() 方法的示例：

范例1：


// Java program to demonstrate
// subList() method for String value
  
import java.util.*;
  
public class GFG1 {
    public static void main(String[] argv)
        throws Exception
    {
  
        try {
  
            // Creating object of AbstractList<Integer>
            AbstractList<String>
                arrlist = new ArrayList<String>();
  
            // Populating arrlist1
            arrlist.add("A");
            arrlist.add("B");
            arrlist.add("C");
            arrlist.add("D");
            arrlist.add("E");
  
            // print arrlist
            System.out.println("Original AbstractList:"
                               + arrlist);
  
            // getting the subList
            // using subList() method
            List<String> arrlist2 = arrlist.subList(2, 4);
  
            // print the subList
            System.out.println("Sublist of AbstractList:"
                               + arrlist2);
        }
  
        catch (IndexOutOfBoundsException e) {
            System.out.println(e);
        }
  
        catch (IllegalArgumentException e) {
            System.out.println(e);
        }
    }
}

输出：

Original AbstractList:[A, B, C, D, E]
Sublist of AbstractList:[C, D]

范例2：对于 IndexOutOfBoundsException


// Java program to demonstrate
// subList() method for IndexOutOfBoundsException
  
import java.util.*;
  
public class GFG1 {
    public static void main(String[] argv)
        throws Exception
    {
  
        try {
  
            // Creating object of AbstractList<Integer>
            AbstractList<String>
                arrlist = new ArrayList<String>();
  
            // Populating arrlist1
            arrlist.add("A");
            arrlist.add("B");
            arrlist.add("C");
            arrlist.add("D");
            arrlist.add("E");
  
            // print arrlist
            System.out.println("Original AbstractList:"
                               + arrlist);
  
            // getting the subList
            // using subList() method
            System.out.println("\nEnd index value is out of range");
            List<String> arrlist2 = arrlist.subList(2, 7);
  
            // print the subList
            System.out.println("Sublist of AbstractList:"
                               + arrlist2);
        }
  
        catch (IndexOutOfBoundsException e) {
            System.out.println(e);
        }
  
        catch (IllegalArgumentException e) {
            System.out.println(e);
        }
    }
}

输出：

Original AbstractList:[A, B, C, D, E]

End index value is out of range
java.lang.IndexOutOfBoundsException:toIndex = 7

范例3：对于 IllegalArgumentException


// Java program to demonstrate
// subList() method for IllegalArgumentException
  
import java.util.*;
  
public class GFG1 {
    public static void main(String[] argv) throws Exception
    {
  
        try {
  
            // Creating object of AbstractList<Integer>
            AbstractList<String>
                arrlist = new ArrayList<String>();
  
            // Populating arrlist1
            arrlist.add("A");
            arrlist.add("B");
            arrlist.add("C");
            arrlist.add("D");
            arrlist.add("E");
  
            // print arrlist
            System.out.println("Original AbstractList:"
                               + arrlist);
  
            // getting the subList
            // using subList() method
            System.out.println("\nEndpoint indices "
                               + "are out of order"
                               + " (fromIndex > toIndex)");
            List<String> arrlist2 = arrlist.subList(7, 2);
  
            // print the subList
            System.out.println("Sublist of AbstractList:"
                               + arrlist2);
        }
  
        catch (IndexOutOfBoundsException e) {
            System.out.println(e);
        }
  
        catch (IllegalArgumentException e) {
            System.out.println(e);
        }
    }
}

输出：

Original AbstractList:[A, B, C, D, E]

Endpoint indices are out of order (fromIndex > toIndex)
java.lang.IllegalArgumentException:fromIndex(7) > toIndex(2)

相关用法

注：本文由纯净天空筛选整理自shubhamsrivastava1490大神的英文原创作品 AbstractList subList() method in Java with Examples。非经特殊声明，原始代码版权归原作者所有，本译文未经允许或授权，请勿转载或复制。