Java 中的抽象列表子列表()方法，示例
原文:https://www . geesforgeks . org/abstract list-sublist-method-in-Java-with-examples/
Java . util . abstract list类的 subList() 方法用于返回此列表中指定的 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。返回的列表支持所有可选的列表操作。
语法:
public List<E> subList(int fromIndex, int toIndex)
参数:该方法将以下参数作为参数。
- fromIndex: 子列表的低端点(含)
- 到索引:子列表的高端点(不包括)
返回值:该方法返回列表中指定范围的视图。
异常:该方法抛出如下异常。
- IndexOutOfBoundsException:如果端点索引值超出范围(从索引大小开始)
- IllegalArgumentException:如果端点索引无序(从索引>到索引)
下面是说明 subList()方法的示例:
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);
}
}
}
Output:
Original AbstractList: [A, B, C, D, E]
Sublist of AbstractList: [C, D]
示例 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);
}
}
}
Output:
Original AbstractList: [A, B, C, D, E]
End index value is out of range
java.lang.IndexOutOfBoundsException: toIndex = 7
示例 3: 适用于非法文档异常
// 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);
}
}
}
Output:
Original AbstractList: [A, B, C, D, E]
Endpoint indices are out of order (fromIndex > toIndex)
java.lang.IllegalArgumentException: fromIndex(7) > toIndex(2)
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