在Java中使用示例Collections.singleton()方法

java.util.Collections.singleton() 方法是java.util.Collections类的方法。它可以创建一个不可变的集合，其中只包含一个指定的元素。该方法的一个应用是从集合中删除元素，如List和Set。

语法：

public static Set singleton(T obj)
and
public static List singletonList(T obj)

参数： obj：要存储在返回的列表或集中的元素。
返回值： 一个包含指定对象的不可变列表/集合。

示例：
我的List列表为{“Geeks”, “code”, “Practice”, ” Error”, “Java”, “Class”, “Error”, “Practice”, “Java” }

使用singleton()方法一次删除所有”Error”元素。
myList.removeAll(Collections.singleton(“Error”));

使用singleton()和removeAll()方法后，我们得到以下结果。
{“Geeks”, “code”, “Practice”, “Java”, “Class”, “Practice”, “Java” }

// Java程序演示singleton()方法的工作原理
import java.util.*;
 
class GFG {
    public static void main(String args[])
    {
        String[] geekslist = { "1", "2", "4", "2", "1", "2",
                               "3", "1", "3", "4", "3", "3" };
 
        // 创建列表并在不使用singleton()的情况下删除元素
        List geekslist1 = new ArrayList(Arrays.asList(geekslist));
        System.out.println("Original geeklist1: " + geekslist1);
 
        geekslist1.remove("1");
        System.out.println("geekslist1 after removal of 1 without"
                           + " singleton " + geekslist1);
        geekslist1.remove("1");
        System.out.println("geekslist1 after removal of 1 without"
                           + " singleton " + geekslist1);
        geekslist1.remove("2");
        System.out.println("geekslist1 after removal of 2 without"
                           + " singleton " + geekslist1);
 
        // 创建另一个列表，并使用singleton()方法删除其元素
        List geekslist2 = new ArrayList(Arrays.asList(geekslist));
        System.out.println("\nOriginal geeklist2: " + geekslist2);
 
        // 从所有出现的"1"中选择性地删除
        geekslist2.removeAll(Collections.singleton("1"));
        System.out.println("geekslist2 after removal of 1 with "
                           + "singleton:" + geekslist2);
 
        // 从所有出现的"4"中选择性地删除
        geekslist2.removeAll(Collections.singleton("4"));
        System.out.println("geekslist2 after removal of 4 with "
                           + "singleton:" + geekslist2);
 
        // 从所有出现的“3”中选择性地删除
        geekslist2.removeAll(Collections.singleton("3"));
        System.out.println("geekslist2 after removal of 3 with"
                           + " singleton: " + geekslist2);
    }
}

输出：

Original geeklist1                 [1, 2, 4, 2, 1, 2, 3, 1, 3, 4, 3, 3]
geekslist1 after removal of 1 without singleton [2, 4, 2, 1, 2, 3, 1, 3, 4, 3, 3]
geekslist1 after removal of 1 without singleton [2, 4, 2, 2, 3, 1, 3, 4, 3, 3]
geekslist1 after removal of 2 without singleton [4, 2, 2, 3, 1, 3, 4, 3, 3]

Original geeklist2                 [1, 2, 4, 2, 1, 2, 3, 1, 3, 4, 3, 3]
geekslist2 after removal of 1 with singleton     [2, 4, 2, 2, 3, 3, 4, 3, 3]
geekslist2 after removal of 4 with singleton     [2, 2, 2, 3, 3, 3, 3]
geekslist2 after removal of 3 with singleton     [2, 2, 2]