我需要创建一个从双向链表中删除给定节点(名为“Jack”的节点)的方法。
这是我的代码:
链表类:
class DoublyLinkedList
{
public Node head, current;
public void AddNode(object n) // add a new node
{
if (head == null)
{
head = new Node(n); //head is pointed to the 1st node in list
current = head;
}
else
{
while (current.next != null)
{
current = current.next;
}
current.next = new Node(n, current); //current is pointed to the newly added node
}
}
public void RemoveNode(object n)
{
}
public String PrintNode() // print nodes
{
String Output = "";
Node printNode = head;
if (printNode != null)
{
while (printNode != null)
{
Output += printNode.data.ToString() + "\r\n";
printNode = printNode.next;
}
}
else
{
Output += "No items in Doubly Linked List";
}
return Output;
}
}
执行按钮代码: 我已经添加了3个节点,我想删除“Jack”节点。
private void btnExecute_Click(object sender, EventArgs e)
{
DoublyLinkedList dll = new DoublyLinkedList();
//add new nodes
dll.AddNode("Tom");
dll.AddNode("Jack");
dll.AddNode("Mary");
//print nodes
txtOutput.Text = dll.PrintNode();
}
答案 0 :(得分:1)
n
n.Next
不是null
,请将n.Next.Prev
设为n.Prev
n.Prev
不是null
,请将n.Prev.Next
设为n.Next
n == head
,请将head
设置为n.Next
基本上,您找到要删除的节点,并使其左侧的节点指向其右侧的节点,反之亦然。
要查找节点n
,您可以执行以下操作:
public bool Remove(object value)
{
Node current = head;
while(current != null && current.Data != value)
current = current.Next;
//value was not found, return false
if(current == null)
return false;
//...
}
注意:这些算法通常涉及两个不变量。您必须确保在任何时候,第一个节点的Prev
属性和最后一个节点的Next
属性为空 - 您可以将其读作:“没有节点出现在第一个节点之前,并且没有节点位于最后一个节点“。
答案 1 :(得分:0)
您的代码应包含Previous
指针,以使其成为双向链接列表。
public void RemoveNode(object n)
{
Node lcurrent = head;
while(lcurrent!=null && lcurrent.Data!=n) //assuming data is an object
{
lcurrent = lcurrent.Next;
}
if(lcurrent != null)
{
if(lcurrent==current) current = current.Previous; //update current
if(lcurrent==head)
{
head = lcurrent.Next;
}
else
{
lcurrent.Previous.Next = lcurrent.Next;
lcurrent.Next.Previous = lcurrent.Previous;
lcurrent.Next = null;
lcurrent.Previous = null;
}
}
}