How to use HashSet.remove() method in Java to remove elements from a collection?

Use the HashSet.remove() method in Java to delete specified elements from the collection. HashSet is a collection class that implements the Set interface. It does not allow the storage of duplicate elements and does not guarantee the order of elements. When operating a HashSet, you can use the remove() method to delete elements in the set.

The remove() method of HashSet has two overloaded forms:

1. boolean remove(Object obj): Remove the specified object from the collection. Returns true if the deletion is successful; false if the specified object does not exist in the collection.
2. boolean removeIf(Predicate super E> filter): Remove elements that meet the condition from the collection. Returns true if at least one element was successfully removed; otherwise returns false.

The following is a sample code that uses the HashSet.remove() method to remove elements from a collection:

import java.util.HashSet; public class HashSetRemoveExample { public static void main(String[] args) { // 创建一个HashSet集合 HashSet set = new HashSet<>(); // 向HashSet集合中添加元素 set.add("Java"); set.add("Python"); set.add("C++"); System.out.println("HashSet集合中的元素：" + set); // 删除指定元素"C++" boolean isRemoved = set.remove("C++"); if (isRemoved) { System.out.println("成功删除元素C++"); System.out.println("删除后的HashSet集合中的元素：" + set); } else { System.out.println("HashSet集合中不存在元素C++"); } // 删除集合中所有以"P"开头的元素 boolean isRemovedAll = set.removeIf(s -> s.startsWith("P")); if (isRemovedAll) { System.out.println("成功删除满足条件的元素"); System.out.println("删除后的HashSet集合中的元素：" + set); } else { System.out.println("HashSet集合中不存在满足条件的元素"); } } }

Run the above code, the output result is:

HashSet集合中的元素：[Java, Python, C++] 成功删除元素C++ 删除后的HashSet集合中的元素：[Java, Python] 成功删除满足条件的元素 删除后的HashSet集合中的元素：[Java]

Okay As you can see, using the HashSet.remove() method can conveniently remove specified elements or elements that meet conditions from the set. Before deleting an element, you need to ensure that the element exists in the collection to avoid errors.

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