Java AbstractList subList()用法及代碼示例

java.util.AbstractList 類的 subList() 方法用於返回此列表在指定 fromIndex(包括)和 toIndex(不包括)之間的部分的視圖。 (如果 fromIndex 和 toIndex 相等，則返回的列表為空。)

返回的列表由該列表支持，因此返回列表中的非結構性更改會反映在該列表中，反之亦然。返回的列表支持所有可選的列表操作。

用法:

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

參數：此方法將以下參數作為參數。

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

返回值：此方法返回此列表中指定範圍的視圖。

異常：此方法引發以下異常。

IndexOutOfBoundsException:如果端點索引值超出範圍(fromIndex 大小)
IllegalArgumentException:如果端點索引亂序(fromIndex > toIndex)

以下是說明 subList() 方法的示例：

範例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);
        }
    }
}

輸出：

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

範例2：對於 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);
        }
    }
}

輸出：

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

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

範例3：對於 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);
        }
    }
}

輸出：

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

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

相關用法

注：本文由純淨天空篩選整理自shubhamsrivastava1490大神的英文原創作品 AbstractList subList() method in Java with Examples。非經特殊聲明，原始代碼版權歸原作者所有，本譯文未經允許或授權，請勿轉載或複製。