C# - LinkedList - 如何删除指定节点后的所有节点? [英] C# - LinkedList - How to remove all nodes after specified node?

问题描述

我正在使用通用 LinkedList 实现撤消/重做缓冲区.

I am implementing an undo/redo buffer with generic LinkedList.

在这种状态下:
[顶部]
state4(撤销)
state3(撤销)
state2 <-- 当前状态
状态1
[底部]

In this state:
[Top]
state4 (undone)
state3 (undone)
state2 <-- current state
state1
[bottom]

当我执行 Push 时，我想删除当前状态之后的所有状态，并推送新的状态.

When I do a Push, I would like to remove all states after the current one, and push the new one.

我目前的绕过是做 while (currentState != list.last), list.removeLast(); 但它很烂

My current bypass is to do while (currentState != list.last), list.removeLast(); but it sucks

LinkedList 只支持 Remove、RemoveFirst &删除最后一个...

LinkedList just support Remove, RemoveFirst & removeLast...

我想要像 RemoveAllNodesAfter(LinkedListNode ...) 之类的东西?

I would like something like RemoveAllNodesAfter(LinkedListNode ...) ?

如何在不遍历所有节点的情况下很好地编码?也许有扩展?...

How can I code that nicely, without iterating throught all nodes ? Maybe with extensions ?...

推荐答案

我在标准 LinkedList<T> 中看不到任何可以让您执行此操作的内容.您可以查看 PowerCollections 和 C5 集合 如果你愿意 - 或者只是滚动你自己的 LinkedList 类型.这是实现起来更简单的集合之一，特别是如果您可以及时"地添加功能.

I can't see anything in the standard LinkedList<T> which lets you do this. You could look in PowerCollections and the C5 collections if you want - or just roll your own LinkedList type. It's one of the simpler collections to implement, especially if you can add functionality in a "just in time" manner.

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