从单链表中删除节点 [英] Remove node from single linked list
问题描述
，你可以这样做：
- 找到要删除的节点
- node.previous.next = node.next
- node.next.previous = node.previous
- node.previous = NULL
- node.next = NULL
- 节点的处置，如果你在
非GC环境是
。
- Find the node to remove.
- node.previous.next = node.next
- node.next.previous = node.previous
- node.previous = null
- node.next = null
- Dispose of node if you're in a non-GC environment
如果您的列表是一个双链接。
If your list is a double linked.
但你如何用一个链表呢？
我已经尝试了很多东西，有没有效果:(
我只是把它删除特定的索引，而不是或它什么都没有。
But how do you do it with a single linked list? I have tried a lot of things, with no avail :( I simply get it to remove a specific index instead or it does nothing at all
推荐答案
开始在列表的开始，维持到当前项目（ CURRENTITEM ）和上一个项目的引用（ previousItem ）。线性搜索要始终删除与 previousItem = CURRENTITEM，CURRENTITEM = currentItem.Next 的项目。如果您要删除的产品列表的头，重新分配名单，以 currentItem.Next 的头部。否则，设置 previousItem。下一步= currentItem.Next 。如果需要的话（如你所说，在非GC环境）处置 CURRENTITEM 的
Start at the beginning of the list. Maintain a reference to the current item (currentItem) and the previous item (previousItem). Linearly search for the item that you want to remove always walking with previousItem = currentItem, currentItem = currentItem.Next. If the item that you want to remove is the head of the list, reassign the head of the list to currentItem.Next. Otherwise, set previousItem.Next = currentItem.Next. If necessary (as you say, in a non-GC environment) dispose of currentItem.
基本上你正在使用 previousItem 来模仿的行为 currentItem.Previous 中，一个双链表的情况下
Basically you are using previousItem to mimic the behavior of a currentItem.Previous in the case of a doubly-linked list.
编辑：这是一个正确实施中删除：
This is a correct implementation of Delete:
public void Delete(int rangeStart, int rangeEnd) {
Node previousNode = null, currentNode = Head;
while (currentNode != null) {
if (currentNode.Data >= rangeStart && currentNode.Data <= rangeEnd) {
if (previousNode == null) {
Initial = currentNode.Next;
}
else {
previousNode.Next = currentNode.Next;
}
}
else {
previousNode = currentNode;
}
currentNode = currentNode.Next;
}
}
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