我正在尝试创建一个链接列表，看看我是否可以，而且我无法理解它。有没有人有一个使用C＃非常简单的链接列表实现的例子？到目前为止，我发现的所有例子都过分夸大了。
答案 0 :(得分：89)
链接列表，其核心是一堆链接在一起的节点。
因此，您需要从一个简单的Node类开始：
public class Node {
public Node next;
public Object data;
}
然后，您的链接列表将作为成员的一个节点表示列表的头部（开头）：
public class LinkedList {
private Node head;
}
然后，您需要通过添加方法向列表添加功能。它们通常涉及所有节点的某种遍历。
public void printAllNodes() {
Node current = head;
while (current != null)
{
Console.WriteLine(current.data);
current = current.next;
}
}
此外，插入新数据是另一种常见操作：
public void Add(Object data) {
Node toAdd = new Node();
toAdd.data = data;
Node current = head;
// traverse all nodes (see the print all nodes method for an example)
current.next = toAdd;
}
这应该是一个很好的起点。
答案 1 :(得分：48)
基于@jjnguy所说的并修复了他的PrintAllNodes（）中的错误，这是完整的Console App示例：
public class Node
{
public Node next;
public Object data;
}
public class LinkedList
{
private Node head;
public void printAllNodes()
{
Node current = head;
while (current != null)
{
Console.WriteLine(current.data);
current = current.next;
}
}
public void AddFirst(Object data)
{
Node toAdd = new Node();
toAdd.data = data;
toAdd.next = head;
head = toAdd;
}
public void AddLast(Object data)
{
if (head == null)
{
head = new Node();
head.data = data;
head.next = null;
}
else
{
Node toAdd = new Node();
toAdd.data = data;
Node current = head;
while (current.next != null)
{
current = current.next;
}
current.next = toAdd;
}
}
}
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Add First:");
LinkedList myList1 = new LinkedList();
myList1.AddFirst("Hello");
myList1.AddFirst("Magical");
myList1.AddFirst("World");
myList1.printAllNodes();
Console.WriteLine();
Console.WriteLine("Add Last:");
LinkedList myList2 = new LinkedList();
myList2.AddLast("Hello");
myList2.AddLast("Magical");
myList2.AddLast("World");
myList2.printAllNodes();
Console.ReadLine();
}
}
答案 2 :(得分：12)
这个很好：
namespace ConsoleApplication1
{
// T is the type of data stored in a particular instance of GenericList.
public class GenericList<T>
{
private class Node
{
// Each node has a reference to the next node in the list.
public Node Next;
// Each node holds a value of type T.
public T Data;
}
// The list is initially empty.
private Node head = null;
// Add a node at the beginning of the list with t as its data value.
public void AddNode(T t)
{
Node newNode = new Node();
newNode.Next = head;
newNode.Data = t;
head = newNode;
}
// The following method returns the data value stored in the last node in
// the list. If the list is empty, the default value for type T is
// returned.
public T GetFirstAdded()
{
// The value of temp is returned as the value of the method.
// The following declaration initializes temp to the appropriate
// default value for type T. The default value is returned if the
// list is empty.
T temp = default(T);
Node current = head;
while (current != null)
{
temp = current.Data;
current = current.Next;
}
return temp;
}
}
}
测试代码：
static void Main(string[] args)
{
// Test with a non-empty list of integers.
GenericList<int> gll = new GenericList<int>();
gll.AddNode(5);
gll.AddNode(4);
gll.AddNode(3);
int intVal = gll.GetFirstAdded();
// The following line displays 5.
System.Console.WriteLine(intVal);
}
我在msdn here
上遇到过它答案 3 :(得分：5)
我是初学者，这对我有所帮助：
class List
{
private Element Root;
}
首先创建包含所有方法的类List。 然后你创建Node-Class，我将其称为Element
class Element
{
public int Value;
public Element Next;
}
然后您可以开始向List类添加方法。例如，这是一个“添加”方法。
public void Add(int value)
{
Element newElement = new Element();
newElement.Value = value;
Element rootCopy = Root;
Root = newElement;
newElement.Next = rootCopy;
Console.WriteLine(newElement.Value);
}
答案 4 :(得分：3)
public class Node
{
private Object data;
public Node next {get;set;}
public Node(Object data)
{
this.data = data;
}
}
public class Linkedlist
{
Node head;
public void Add(Node n)
{
n.Next = this.Head;
this.Head = n;
}
}
使用：
LinkedList sample = new LinkedList();
sample.add(new Node("first"));
sample.Add(new Node("second"))
答案 5 :(得分：3)
这里有一个IEnumerable和一个递归反向方法虽然它不比Reverse方法中的while循环快，但它们都是O（n）：
public class LinkedList<T> : IEnumerable
{
private Node<T> _head = null;
public Node<T> Add(T value)
{
var node = new Node<T> {Value = value};
if (_head == null)
{
_head = node;
}
else
{
var current = _head;
while (current.Next != null)
{
current = current.Next;
}
current.Next = node; //new head
}
return node;
}
public T Remove(Node<T> node)
{
if (_head == null)
return node.Value;
if (_head == node)
{
_head = _head.Next;
node.Next = null;
return node.Value;
}
var current = _head;
while (current.Next != null)
{
if (current.Next == node)
{
current.Next = node.Next;
return node.Value;
}
current = current.Next;
}
return node.Value;
}
public void Reverse()
{
Node<T> prev = null;
var current = _head;
if (current == null)
return;
while (current != null)
{
var next = current.Next;
current.Next = prev;
prev = current;
current = next;
}
_head = prev;
}
public void ReverseRecurisve()
{
reverseRecurive(_head, null);
}
private void reverseRecurive(Node<T> current, Node<T> prev)
{
if (current.Next == null)
{
_head = current;
_head.Next = prev;
return;
}
var next = current.Next;
current.Next = prev;
reverseRecurive(next, current);
}
public IEnumerator<T> Enumerator()
{
var current = _head;
while (current != null)
{
yield return current.Value;
current = current.Next;
}
}
public IEnumerator GetEnumerator()
{
return Enumerator();
}
}
public class Node<T>
{
public T Value { get; set; }
public Node<T> Next { get; set; }
}
答案 6 :(得分：2)
这是一个很好的实现。
该列表可以包含任何数据类型。
using System;
class Node<Type> : LinkedList<Type>
{ // Why inherit from LinkedList? A: We need to use polymorphism.
public Type value;
public Node(Type value) { this.value = value; }
}
class LinkedList<Type>
{
Node<Type> next; // This member is treated as head in class LinkedList, but treated as next element in class Node.
/// <summary> if x is in list, return previos pointer of x. (We can see any class variable as a pointer.)
/// if not found, return the tail of the list. </summary>
protected LinkedList<Type> Previos(Type x)
{
LinkedList<Type> p = this; // point to head
for (; p.next != null; p = p.next)
if (p.next.value.Equals(x))
return p; // find x, return the previos pointer.
return p; // not found, p is the tail.
}
/// <summary> return value: true = success ; false = x not exist </summary>
public bool Contain(Type x) { return Previos(x).next != null ? true : false; }
/// <summary> return value: true = success ; false = fail to add. Because x already exist.
/// </summary> // why return value? If caller want to know the result, they don't need to call Contain(x) before, the action waste time.
public bool Add(Type x)
{
LinkedList<Type> p = Previos(x);
if (p.next != null) // Find x already in list
return false;
p.next = new Node<Type>(x);
return true;
}
/// <summary> return value: true = success ; false = x not exist </summary>
public bool Delete(Type x)
{
LinkedList<Type> p = Previos(x);
if (p.next == null)
return false;
//Node<Type> node = p.next;
p.next = p.next.next;
//node.Dispose(); // GC dispose automatically.
return true;
}
public void Print()
{
Console.Write("List: ");
for (Node<Type> node = next; node != null; node = node.next)
Console.Write(node.value.ToString() + " ");
Console.WriteLine();
}
}
class Test
{
static void Main()
{
LinkedList<int> LL = new LinkedList<int>();
if (!LL.Contain(0)) // Empty list
Console.WriteLine("0 is not exist.");
LL.Print();
LL.Add(0); // Add to empty list
LL.Add(1); LL.Add(2); // attach to tail
LL.Add(2); // duplicate add, 2 is tail.
if (LL.Contain(0))// Find existed element which is head
Console.WriteLine("0 is exist.");
LL.Print();
LL.Delete(0); // Delete head
LL.Delete(2); // Delete tail
if (!LL.Delete(0)) // Delete non-exist element
Console.WriteLine("0 is not exist.");
LL.Print();
Console.ReadLine();
}
}
答案 7 :(得分：2)
我将从约瑟夫·阿尔巴哈里和本·阿尔巴哈里的“果壳中的C＃6.0”一书中摘录出来。
以下是使用LinkedList的演示：
var tune = new LinkedList<string>();
tune.AddFirst ("do"); // do
tune.AddLast ("so"); // do - so
tune.AddAfter (tune.First, "re"); // do - re- so
tune.AddAfter (tune.First.Next, "mi"); // do - re - mi- so
tune.AddBefore (tune.Last, "fa"); // do - re - mi - fa- so
tune.RemoveFirst(); // re - mi - fa - so
tune.RemoveLast(); // re - mi - fa
LinkedListNode<string> miNode = tune.Find ("mi");
tune.Remove (miNode); // re - fa
tune.AddFirst (miNode); // mi- re - fa
foreach (string s in tune) Console.WriteLine (s);
答案 8 :(得分：1)
public class Node<T>
{
public T item;
public Node<T> next;
public Node()
{
this.next = null;
}
}
class LinkList<T>
{
public Node<T> head { get; set; }
public LinkList()
{
this.head = null;
}
public void AddAtHead(T item)
{
Node<T> newNode = new Node<T>();
newNode.item = item;
if (this.head == null)
{
this.head = newNode;
}
else
{
newNode.next = head;
this.head = newNode;
}
}
public void AddAtTail(T item)
{
Node<T> newNode = new Node<T>();
newNode.item = item;
if (this.head == null)
{
this.head = newNode;
}
else
{
Node<T> temp = this.head;
while (temp.next != null)
{
temp = temp.next;
}
temp.next = newNode;
}
}
public void DeleteNode(T item)
{
if (this.head.item.Equals(item))
{
head = head.next;
}
else
{
Node<T> temp = head;
Node<T> tempPre = head;
bool matched = false;
while (!(matched = temp.item.Equals(item)) && temp.next != null)
{
tempPre = temp;
temp = temp.next;
}
if (matched)
{
tempPre.next = temp.next;
}
else
{
Console.WriteLine("Value not found!");
}
}
}
public bool searchNode(T item)
{
Node<T> temp = this.head;
bool matched = false;
while (!(matched = temp.item.Equals(item)) && temp.next != null)
{
temp = temp.next;
}
return matched;
}
public void DisplayList()
{
Console.WriteLine("Displaying List!");
Node<T> temp = this.head;
while (temp != null)
{
Console.WriteLine(temp.item);
temp = temp.next;
}
}
}
答案 9 :(得分：0)
public class DynamicLinkedList
{
private class Node
{
private object element;
private Node next;
public object Element
{
get { return this.element; }
set { this.element = value; }
}
public Node Next
{
get { return this.next; }
set { this.next = value; }
}
public Node(object element, Node prevNode)
{
this.element = element;
prevNode.next = this;
}
public Node(object element)
{
this.element = element;
next = null;
}
}
private Node head;
private Node tail;
private int count;
public DynamicLinkedList()
{
this.head = null;
this.tail = null;
this.count = 0;
}
public void AddAtLastPosition(object element)
{
if (head == null)
{
head = new Node(element);
tail = head;
}
else
{
Node newNode = new Node(element, tail);
tail = newNode;
}
count++;
}
public object GetLastElement()
{
object lastElement = null;
Node currentNode = head;
while (currentNode != null)
{
lastElement = currentNode.Element;
currentNode = currentNode.Next;
}
return lastElement;
}
}
测试：
static void Main(string[] args)
{
DynamicLinkedList list = new DynamicLinkedList();
list.AddAtLastPosition(1);
list.AddAtLastPosition(2);
list.AddAtLastPosition(3);
list.AddAtLastPosition(4);
list.AddAtLastPosition(5);
object lastElement = list.GetLastElement();
Console.WriteLine(lastElement);
}
答案 10 :(得分：0)
Dmytro做得很好，但这是一个更简洁的版本。
class Program
{
static void Main(string[] args)
{
LinkedList linkedList = new LinkedList(1);
linkedList.Add(2);
linkedList.Add(3);
linkedList.Add(4);
linkedList.AddFirst(0);
linkedList.Print();
}
}
public class Node
{
public Node(Node next, Object value)
{
this.next = next;
this.value = value;
}
public Node next;
public Object value;
}
public class LinkedList
{
public Node head;
public LinkedList(Object initial)
{
head = new Node(null, initial);
}
public void AddFirst(Object value)
{
head = new Node(head, value);
}
public void Add(Object value)
{
Node current = head;
while (current.next != null)
{
current = current.next;
}
current.next = new Node(null, value);
}
public void Print()
{
Node current = head;
while (current != null)
{
Console.WriteLine(current.value);
current = current.next;
}
}
}
答案 11 :(得分：0)
简单的c＃程序，用于实现具有 AddItemStart，AddItemEnd，RemoveItemStart，RemoveItemEnd和DisplayAllItems
操作的单链接列表 using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace SingleLinkedList
{
class Program
{
Node head;
Node current;
int counter = 0;
public Program()
{
head = new Node();
current = head;
}
public void AddStart(object data)
{
Node newnode = new Node();
newnode.next = head.next;
newnode.data = data;
head.next = newnode;
counter++;
}
public void AddEnd(object data)
{
Node newnode = new Node();
newnode.data = data;
current.next = newnode;
current = newnode;
counter++;
}
public void RemoveStart()
{
if (counter > 0)
{
head.next = head.next.next;
counter--;
}
else
{
Console.WriteLine("No element exist in this linked list.");
}
}
public void RemoveEnd()
{
if (counter > 0)
{
Node prevNode = new Node();
Node cur = head;
while (cur.next != null)
{
prevNode = cur;
cur = cur.next;
}
prevNode.next = null;
}
else
{
Console.WriteLine("No element exist in this linked list.");
}
}
public void Display()
{
Console.Write("Head ->");
Node curr = head;
while (curr.next != null)
{
curr = curr.next;
Console.WriteLine(curr.data.ToString());
}
}
public class Node
{
public object data;
public Node next;
}
static void Main(string[] args)
{
Program p = new Program();
p.AddEnd(2);
p.AddStart(1);
p.AddStart(0);
p.AddEnd(3);
p.Display();
p.RemoveStart();
Console.WriteLine("Removed node from Start");
p.Display();
Console.WriteLine("Removed node from End");
p.RemoveEnd();
p.Display();
Console.ReadKey();
}
}
}
答案 12 :(得分：0)
所选答案没有迭代器;它更基本，但也许不那么有用。
这是一个带有迭代器/枚举器的程序。我的实施是基于Sedgewick的包;见http://algs4.cs.princeton.edu/13stacks/Bag.java.html
void Main()
{
var b = new Bag<string>();
b.Add("bike");
b.Add("erasmus");
b.Add("kumquat");
b.Add("beaver");
b.Add("racecar");
b.Add("barnacle");
foreach (var thing in b)
{
Console.WriteLine(thing);
}
}
// Define other methods and classes here
public class Bag<T> : IEnumerable<T>
{
public Node<T> first;// first node in list
public class Node<T>
{
public T item;
public Node<T> next;
public Node(T item)
{
this.item = item;
}
}
public void Add(T item)
{
Node<T> oldFirst = first;
first = new Node<T>(item);
first.next = oldFirst;
}
IEnumerator IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
public IEnumerator<T> GetEnumerator()
{
return new BagEnumerator<T>(this);
}
public class BagEnumerator<V> : IEnumerator<T>
{
private Node<T> _head;
private Bag<T> _bag;
private Node<T> _curNode;
public BagEnumerator(Bag<T> bag)
{
_bag = bag;
_head = bag.first;
_curNode = default(Node<T>);
}
public T Current
{
get { return _curNode.item; }
}
object IEnumerator.Current
{
get { return Current; }
}
public bool MoveNext()
{
if (_curNode == null)
{
_curNode = _head;
if (_curNode == null)
return false;
return true;
}
if (_curNode.next == null)
return false;
else
{
_curNode = _curNode.next;
return true;
}
}
public void Reset()
{
_curNode = default(Node<T>); ;
}
public void Dispose()
{
}
}
}
答案 13 :(得分：0)
我创建了以下具有许多功能的LinkedList代码。它可以在CodeBase github public repo下公开使用。
课程：
Node和LinkedList
字母和字母设置器： First和Last
功能：
AddFirst(data), AddFirst(node), AddLast(data), RemoveLast(), AddAfter(node, data), RemoveBefore(node), Find(node), Remove(foundNode), Print(LinkedList)
using System;
using System.Collections.Generic;
namespace Codebase
{
public class Node
{
public object Data { get; set; }
public Node Next { get; set; }
public Node()
{
}
public Node(object Data, Node Next = null)
{
this.Data = Data;
this.Next = Next;
}
}
public class LinkedList
{
private Node Head;
public Node First
{
get => Head;
set
{
First.Data = value.Data;
First.Next = value.Next;
}
}
public Node Last
{
get
{
Node p = Head;
//Based partially on https://en.wikipedia.org/wiki/Linked_list
while (p.Next != null)
p = p.Next; //traverse the list until p is the last node.The last node always points to NULL.
return p;
}
set
{
Last.Data = value.Data;
Last.Next = value.Next;
}
}
public void AddFirst(Object data, bool verbose = true)
{
Head = new Node(data, Head);
if (verbose) Print();
}
public void AddFirst(Node node, bool verbose = true)
{
node.Next = Head;
Head = node;
if (verbose) Print();
}
public void AddLast(Object data, bool Verbose = true)
{
Last.Next = new Node(data);
if (Verbose) Print();
}
public Node RemoveFirst(bool verbose = true)
{
Node temp = First;
Head = First.Next;
if (verbose) Print();
return temp;
}
public Node RemoveLast(bool verbose = true)
{
Node p = Head;
Node temp = Last;
while (p.Next != temp)
p = p.Next;
p.Next = null;
if (verbose) Print();
return temp;
}
public void AddAfter(Node node, object data, bool verbose = true)
{
Node temp = new Node(data);
temp.Next = node.Next;
node.Next = temp;
if (verbose) Print();
}
public void AddBefore(Node node, object data, bool verbose = true)
{
Node temp = new Node(data);
Node p = Head;
while (p.Next != node) //Finding the node before
{
p = p.Next;
}
temp.Next = p.Next; //same as = node
p.Next = temp;
if (verbose) Print();
}
public Node Find(object data)
{
Node p = Head;
while (p != null)
{
if (p.Data == data)
return p;
p = p.Next;
}
return null;
}
public void Remove(Node node, bool verbose = true)
{
Node p = Head;
while (p.Next != node)
{
p = p.Next;
}
p.Next = node.Next;
if (verbose) Print();
}
public void Print()
{
Node p = Head;
while (p != null) //LinkedList iterator
{
Console.Write(p.Data + " ");
p = p.Next; //traverse the list until p is the last node.The last node always points to NULL.
}
Console.WriteLine();
}
}
}
当她使用Microsoft内置的LinkedList和LinkedListNode回答问题时，使用@yogihosting答案，您可以获得相同的结果：
using System;
using System.Collections.Generic;
using Codebase;
namespace Cmd
{
static class Program
{
static void Main(string[] args)
{
var tune = new LinkedList(); //Using custom code instead of the built-in LinkedList<T>
tune.AddFirst("do"); // do
tune.AddLast("so"); // do - so
tune.AddAfter(tune.First, "re"); // do - re- so
tune.AddAfter(tune.First.Next, "mi"); // do - re - mi- so
tune.AddBefore(tune.Last, "fa"); // do - re - mi - fa- so
tune.RemoveFirst(); // re - mi - fa - so
tune.RemoveLast(); // re - mi - fa
Node miNode = tune.Find("mi"); //Using custom code instead of the built in LinkedListNode
tune.Remove(miNode); // re - fa
tune.AddFirst(miNode); // mi- re - fa
}
}
答案 14 :(得分：-1)
链表是基于节点的数据结构。每个节点设计有两个部分（数据和节点引用）。实际上，数据总是存储在数据部分（也许是原始数据类型，例如Int，Float等，或者我们也可以存储用户定义的数据类型，例如对象引用），并且类似节点引用还应该包含对下一个节点的引用，如果没有下一个节点，那么链将结束。
此链将继续直到没有下一个节点参考点的任何节点。
请从我的技术博客中找到源代码-http://www.algonuts.info/linked-list-program-in-java.html
package info.algonuts;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.List;
class LLNode {
int nodeValue;
LLNode childNode;
public LLNode(int nodeValue) {
this.nodeValue = nodeValue;
this.childNode = null;
}
}
class LLCompute {
private static LLNode temp;
private static LLNode previousNode;
private static LLNode newNode;
private static LLNode headNode;
public static void add(int nodeValue) {
newNode = new LLNode(nodeValue);
temp = headNode;
previousNode = temp;
if(temp != null)
{ compute(); }
else
{ headNode = newNode; } //Set headNode
}
private static void compute() {
if(newNode.nodeValue < temp.nodeValue) { //Sorting - Ascending Order
newNode.childNode = temp;
if(temp == headNode)
{ headNode = newNode; }
else if(previousNode != null)
{ previousNode.childNode = newNode; }
}
else
{
if(temp.childNode == null)
{ temp.childNode = newNode; }
else
{
previousNode = temp;
temp = temp.childNode;
compute();
}
}
}
public static void display() {
temp = headNode;
while(temp != null) {
System.out.print(temp.nodeValue+" ");
temp = temp.childNode;
}
}
}
public class LinkedList {
//Entry Point
public static void main(String[] args) {
//First Set Input Values
List <Integer> firstIntList = new ArrayList <Integer>(Arrays.asList(50,20,59,78,90,3,20,40,98));
Iterator<Integer> ptr = firstIntList.iterator();
while(ptr.hasNext())
{ LLCompute.add(ptr.next()); }
System.out.println("Sort with first Set Values");
LLCompute.display();
System.out.println("\n");
//Second Set Input Values
List <Integer> secondIntList = new ArrayList <Integer>(Arrays.asList(1,5,8,100,91));
ptr = secondIntList.iterator();
while(ptr.hasNext())
{ LLCompute.add(ptr.next()); }
System.out.println("Sort with first & Second Set Values");
LLCompute.display();
System.out.println();
}
}
答案 15 :(得分：-1)
我有一个双向链表，可以用作堆栈或队列。如果您看一下代码并考虑它的功能以及如何执行，我敢打赌，您将了解它的所有内容。很抱歉，但是不知何故我无法在此处添加完整代码，所以我在这里是链表的链接（解决方案中也有二叉树）：https://github.com/szabeast/LinkedList_and_BinaryTree