[英]reverse a linked list in a recursive function c#
I'm having problems trying to write a reverse recursive method for a LinkedList class I created in C#. 我在尝试为在C#中创建的LinkedList类编写反向递归方法时遇到问题。 The LinkedList has 2 pointers in it one for the head and the other for the tail:
LinkedList中有2个指针,一个指向头部,另一个指向尾部:
public class Node
{
public object data;
public Node next;
public Node(object Data)
{
this.data = Data;
}
}
public class LinkedList
{
Node head;
Node tail;
public void Add(Node n)
{
if (head == null)
{
head = n;
tail = head;
}
else
{
tail.next = n;
tail = tail.next;
}
}
Now, the recursive reverse function goes like this: 现在,递归反向函数如下所示:
public void reverse_recursive()
{
Node temp_head = head;
if (temp_head == tail)
{
return;
}
while (temp_head != null)
{
if (temp_head.next == tail)
{
tail.next = temp_head;
tail = temp_head;
reverse_recursive();
}
temp_head = temp_head.next;
}
}
I'm having 2 troubles with it: first, a logic problem, I know that head doesn't point to the first node after the reverse. 我遇到了2个麻烦:首先是逻辑问题,我知道头并没有指向反向之后的第一个节点。 The second problem is that i probably do something wrong with the null pointer so the program crashes.
第二个问题是我可能会对null指针做些错误,因此程序崩溃了。
I also give you the main program: 我也给你主程序:
class Program
{
static void Main(string[] args)
{
LinkedList L = new LinkedList();
L.Add(new Node("first"));
L.Add(new Node("second"));
L.Add(new Node("third"));
L.Add(new Node("forth"));
L.PrintNodes();
L.reverse_recursive();
L.PrintNodes();
Console.ReadLine();
}
}
Thank you for helping!! 感谢您的帮助!!
public void Reverse()
{
this.Reverse(this.head);
}
private void Reverse(Node node)
{
if (node != null && node.next != null)
{
// Create temporary references to the nodes,
// because we will be overwriting the lists references.
Node next = node.next;
Node afterNext = node.next.next;
Node currentHead = this.head;
// Set the head to whatever node is next from the current node.
this.head = next;
// Reset the next node for the new head to be the previous head.
this.head.next = currentHead;
// Set the current nodes next node to be the previous next nodes next node :)
node.next = afterNext;
// Keep on trucking.
this.Reverse(node);
}
else
{
this.tail = node;
}
}
public void reverse()
{
reverse_recursive(tail);
Node tmp = tail;
tail = head;
head = tmp;
}
public void reverse_recursive(Node endNode)
{
Node temp_head = head;
if (temp_head == endNode)
{
return;
}
while (temp_head != null)
{
if (temp_head.next == endNode)
{
break;
}
temp_head = temp_head.next;
}
endNode.next = temp_head;
temp_head.next = null;
reverse_recursive(temp_head);
}
Another option over here. 这里的另一个选择。
class Program{
static void Main(string[] args)
{
LinkedList L = new LinkedList();
L.Add(new Node("first"));
L.Add(new Node("second"));
L.Add(new Node("third"));
L.Add(new Node("forth"));
L.PrintNodes();
L.reverse_recursive();
Console.WriteLine("---------------------");
L.PrintNodes();
Console.ReadLine();
}
}
public class Node
{
public object data;
public Node next;
public Node(object Data)
{
this.data = Data;
}
}
public class LinkedList
{
Node head;
Node tail;
public void Add(Node n)
{
if (head == null)
{
head = n;
tail = head;
}
else
{
tail.next = n;
tail = tail.next;
}
}
public void PrintNodes()
{
Node temp = head;
while (temp != null)
{
Console.WriteLine(temp.data);
temp = temp.next;
}
}
private LinkedList p_reverse_recursive(Node first)
{
LinkedList ret;
if (first.next == null)
{
Node aux = createNode(first.data);
ret = new LinkedList();
ret.Add(aux);
return ret;
}
else
{
ret = p_reverse_recursive(first.next);
ret.Add(createNode(first.data));
return ret;
}
}
private Node createNode(Object data)
{
Node node = new Node(data);
return node;
}
public void reverse_recursive()
{
if (head != null)
{
LinkedList aux = p_reverse_recursive(head);
head = aux.head;
tail = aux.tail;
}
}
}
Hope it helps 希望能帮助到你
A second variant 第二种变体
private void p_reverse_recursive2(Node node)
{
if (node != null)
{
Node aux = node.next;
node.next = null;
p_reverse_recursive2(aux);
if (aux != null)
aux.next = node;
}
}
public void reverse_recursive()
{
if (head != null)
{
Node aux = head;
head = tail;
tail = aux;
p_reverse_recursive2(tail);
}
}
Variation on a theme... 主题变化...
public Node Reverse(Node head)
{
if(head == null)
{
return null;
}
Node reversedHead = null;
ReverseHelper(head, out reversedHead);
return reversedHead;
}
public Node ReverseHelper(Node n, out Node reversedHead)
{
if(n.Next == null)
{
reversedHead = n;
return n;
}
var reversedTail = ReverseHelper(n.Next, out reversedHead);
reversedTail.Next = n;
n.Next = null;
return n;
}
}
I was just playing with similar brain teaser with the only difference that LinkedList class only has a definition for head and all the rest of the nodes are linked there. 我只是在玩类似的脑筋急转弯,唯一的区别是LinkedList类仅具有head定义,所有其余节点都链接在那里。 So here is my quick and dirty recursive solution:
所以这是我快速又肮脏的递归解决方案:
public Node ReverseRecursive(Node root)
{
Node temp = root;
if (root.next == null)
return root;
else
root = ReverseRecursive(root.next);
temp.next = null;
Node tail = root.next;
if (tail == null)
root.next = temp;
else
while (tail != null)
{
if (tail.next == null)
{
tail.next = temp;
break;
}
else
tail = tail.next;
}
return root;
}
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