如何从Java集合中弹出项目？

How to pop items from a collection in Java?

提问by Tom Brito

Is there a method in JDK or apache commons to "pop" a list of elements from a java.util.List? I mean, remove the list of elements and return it, like this method:

JDK 或 apache commons 中是否有一种方法可以从 java.util.List 中“弹出”元素列表？我的意思是，删除元素列表并返回它，就像这个方法：

public Collection pop(Collection elementsToPop, Collection elements) {

  Collection popped = new ArrayList();

  for (Object object : elementsToPop) {
    if (elements.contains(object)) {
      elements.remove(object);
      popped.add(object);
    }
  }

  return popped;
}

采纳答案by Mark Peters

If you're looking for a stack-like structure I suggest accepting a Deque(LinkedListis the most common implementation) instead of a Collection.

如果您正在寻找类似堆栈的结构，我建议接受 a Deque(LinkedList是最常见的实现) 而不是Collection.

If you don't actually need to treat it as a stack, just get an iterator from the Collectionand use the remove()method:

如果您实际上不需要将其视为堆栈，只需从 中获取迭代器Collection并使用该remove()方法：

for (Iterator<SomeType> it = elements.iterator(); it.hasNext(); ) {
    SomeType e = it.next();
    it.remove();
    popped.add(e);
}

Do note that remove is an optional operation, and some implementations may throw an UnsupportedOperationException(for example, the iterator returned by a Collection from Collections.unmodifiable...()will).

请注意，remove 是一个可选操作，某些实现可能会抛出一个UnsupportedOperationException（例如，来自Collections.unmodifiable...()will的 Collection 返回的迭代器）。

Edit: After looking more closely at your question, I think you just need this:

编辑：在更仔细地查看您的问题后，我认为您只需要这个：

elements.removeAll(elementsToRemove);

If your main point is you need to know exactly whichelements were actually popped, I think you're stuck with your original code.

如果您的主要观点是您需要确切地知道实际弹出了哪些元素，我认为您被困在原始代码中。

回答by Nikita Rybak

I guess no, because you definition of 'pop' operation is highly non-standard. Usually it takes no arguments (except collection itself) and returns and removes the top-most one.

我想不会，因为您对“流行”操作的定义是高度非标准的。通常它不接受任何参数（集合本身除外）并返回并删除最顶层的参数。

But once you noted apache commons, this would achieve the same effect as your code.

但是一旦您注意到 apache commons，这将达到与您的代码相同的效果。

Collection result = CollectionUtils.intersection(a, b);
a.removeAll(b);

edit
http://commons.apache.org/collections/api-release/index.html

编辑
http://commons.apache.org/collections/api-release/index.html

回答by mikera

There isn't a method exactly like what you are asking for, but it looks like you are already pretty close with your code.

没有完全符合您要求的方法，但看起来您已经非常接近您的代码。

Some suggestions:

一些建议：

• Consider using removeAll(object) instead of remove(object) if elements is an arbitrary collection since you may need to remove duplicates e.g. if elements is a list.

• contains() is slow for some collection types (e.g. lists) since it needs to traverse the entire data structure. Given that this is in your inner loop you are at risk of O(n^2) performance issues. If you can make the algorithm work with a HashSet or HashMap then contains() will by O(1) and your algorithm will be much more efficient.

• 如果元素是任意集合，请考虑使用 removeAll(object) 而不是 remove(object)，因为您可能需要删除重复项，例如，如果元素是列表。

• contains() 对于某些集合类型（例如列表）来说很慢，因为它需要遍历整个数据结构。鉴于这是在您的内部循环中，您面临 O(n^2) 性能问题的风险。如果您可以使算法与 HashSet 或 HashMap 一起工作，那么 contains() 将通过 O(1) 并且您的算法将效率更高。

回答by Pops

There is no such method in the standard JDK-provided methods. Apache Commons provides the ListUtils.subtract()method.

在标准的 JDK 提供的方法中没有这样的方法。Apache Commons 提供了该ListUtils.subtract()方法。

Edit: As other answerers have noted, your use of the term popis nonstandard. Usually,

编辑：正如其他回答者所指出的，您对该术语的使用pop是非标准的。通常，

The pop operation removes an item from the top of [a stack]

pop 操作从 [a stack] 的顶部移除一个项目

Wikipedia has a nice description of stacks.

维基百科对 stacks有很好的描述。

回答by Alexander M

Linked List provides the functionality as you require, provides a push and pop method.

链表提供您需要的功能，提供推送和弹出方法。

Refer to the documentationas provided:

请参阅提供的文档：