我写了一种创建链表副本的方法。你们能想到比这更好的方法吗?
public static Node Duplicate(Node n)
{
Stack<Node> s = new Stack<Node>();
while (n != null)
{
Node n2 = new Node();
n2.Data = n.Data;
s.Push(n2);
n = n.Next;
}
Node temp = null;
while (s.Count > 0)
{
Node n3 = s.Pop();
n3.Next = temp;
temp = n3;
}
return temp;
}
您可以一口气完成,就像这样:
public static Node Duplicate(Node n)
{
// handle the degenerate case of an empty list
if (n == null) {
return null;
}
// create the head node, keeping it for later return
Node first = new Node();
first.Data = n.Data;
// the 'temp' pointer points to the current "last" node in the new list
Node temp = first;
n = n.Next;
while (n != null)
{
Node n2 = new Node();
n2.Data = n.Data;
// modify the Next pointer of the last node to point to the new last node
temp.Next = n2;
temp = n2;
n = n.Next;
}
return first;
}
@ Greg,我拿走了您的代码,并使其变得更短:)
public static Node Duplicate(Node n)
{
// Handle the degenerate case of an empty list
if (n == null) return null;
// Create the head node, keeping it for later return
Node first = new Node();
Node current = first;
do
{
// Copy the data of the Node
current.Data = n.Data;
current = (current.Next = new Node());
n = n.Next;
} while (n != null)
return first;
}
While结构通常被遗忘,但非常适合这里。一个Node.Clone()方法也很好。
+ 1为Greg提供了很好的示例。
中小型列表的递归方法。
public static Node Duplicate(Node n)
{
if (n == null)
return null;
return new Node() {
Data = n.Data,
Next = Duplicate(n.Next)
};
}
对不起,如果我错过了什么,但是出了什么问题
LinkedList<MyType> clone = new LinkedList<MyType>(originalLinkedList);