反向链表c#

Question

我用 C# 制作了一个 link 列表程序，但我还想反转 link 列表中的数字。该程序运行并让我将数字添加到列表中，但是一旦我添加了数字，数字就不会出现在反向部分，它只会输出“反向列表”。如何以相反的顺序显示数字？

using System;

namespace LinkedList
{
class Program
{
    public class Node
    {
        public int data;
        public Node next;
    };

    static Node add(Node head, int data) 
    {
        Node temp = new Node();
        Node current;
        temp.data = data;
        temp.next = null; 

        if (head == null) 
            head = temp;
        else
        {
            current = head;
            while (current.next != null)
                current = current.next;
            current.next = temp; 
        }
        return head;
    }

     static void reverse_list(Node head)
    {
        Node prev = null, current = head, next = null;

        while (current != null)
            next = current.next;
        current.next = prev;
        prev = current;
        current = next;
    }

    static void print_numbers(Node head)
    {
        while (head != null)
        {
            Console.Write(head.data + " "); 
            head = head.next;
        }
    }

    static Node List(int[] a, int n)
    {
        Node head = null; 
        for (int i = 1; i <= n; i++)
            head = add(head, a[i]);
        return head;
    }

    public static void Main(String[] args)
    {
        
        int n = 10; 
        int[] a;
        a = new int[n + 1];
        a[0] = 0;

        Console.WriteLine("Add values to the list");
        for (int i = 1; i <= n; i++) 
            
            a[i] = int.Parse(Console.ReadLine()); 

        Node head = List(a, n);
        Console.WriteLine("Linked List: ");
        print_numbers(head);
        Console.ReadLine();
        Console.WriteLine();
        Console.WriteLine("Reversed list: ");
        reverse_list(head);
        print_numbers(head);
     }

    }
   }

Answer 1

如果您想对单链表执行此操作，您将不得不抓取该列表和re-link所有内容

对于列表A>B>C>D>E，您需要当前为 B，上一个为 A，下一个为 C。您需要将当前更改为指向上一个，然后将所有内容向前推进一个，直到到达结束

/*
A>B>C>D>E
A<B C>D>E
A<B<C D>E
A<B<C<D<E
*/


var prev = head;      //A
var curr = head.Next; //B
var next = curr.Next; //C, we need this so C is not lost when we re-point B at A in loop line 1

prev.Next = null;     //null<A    B>C>D>E

while(true)
{
  curr.Next = prev;   //null<A<B  C>D>E
  prev = curr;        //prev is now B, was A
  curr = next;        //curr is now C, was B

  if(curr.Next != null){
    next = curr.Next;   //next is now D, was C 
  } else {
    head = curr;
    break;
  }
}

我认为这是逻辑；还没有测试过

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