Java 中的 ConcurrentSkipListSet remove()方法
原文:https://www . geeksforgeeks . org/concurrentskiplistset-remove-method-in-Java/
java . util . concurrentSkiplistset . remove()方法是 Java 中的一个内置函数，如果元素存在于这个集合中，它将被用来移除该元素。
语法:
ConcurrentSkipListSet.remove(Object o)
参数:该功能接受单个参数 o ，即要移除的对象。
返回值:该函数在成功移除对象时返回真布尔值，否则返回假。
下面的程序说明了 ConcurrentSkipListSet.remove()方法:
程序 1: 集合中存在要移除的元素。
// Java Program Demonstrate remove()
// method of ConcurrentSkipListSet
import java.util.concurrent.ConcurrentSkipListSet;
class ConcurrentSkipListSetRemoveExample1 {
public static void main(String[] args)
{
// Initializing the set
ConcurrentSkipListSet<Integer> set =
new ConcurrentSkipListSet<Integer>();
// Adding elements to this set
for (int i = 1; i <= 5; i++)
set.add(i);
// Printing the elements of the set
System.out.println("The elements in the set are:");
for (Integer i : set)
System.out.print(i + " ");
// remove() method will remove the specified
// element from the set
set.remove(1);
set.remove(5);
// Printing the elements of the set
System.out.println("\nRemaining elements in set : ");
for (Integer i : set)
System.out.print(i + " ");
}
}
输出:
The elements in the set are:
1 2 3 4 5
Remaining elements in set :
2 3 4
程序 2: 要移除的元素不在集合中。
// Java Program Demonstrate remove()
// method of ConcurrentSkipListSet
import java.util.concurrent.ConcurrentSkipListSet;
class ConcurrentSkipListSetRemoveExample2 {
public static void main(String[] args)
{
// Initializing the set
ConcurrentSkipListSet<Integer> set =
new ConcurrentSkipListSet<Integer>();
// Adding elements to this set
for (int i = 10; i <= 15; i++)
set.add(i);
// Printing the elements of the set
System.out.println("The elements in the set are:");
for (Integer i : set)
System.out.print(i + " ");
// remove() method will remove the specified
// element from the set
set.remove(1);
set.remove(5);
// Printing the elements of the set
System.out.println("\nRemaining elements in set : ");
for (Integer i : set)
System.out.print(i + " ");
}
}
输出:
The elements in the set are:
10 11 12 13 14 15
Remaining elements in set :
10 11 12 13 14 15
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