C# Anagram Checker 与 LinkedList
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【中文标题】C# Anagram Checker 与 LinkedList【英文标题】：C# Anagram Checker With LinkedList 【发布时间】：2016-09-05 08:53:57 【问题描述】：我正在尝试检查两个单词是否是字谜，并尝试使用 LinkedList 执行此操作。为此，首先，我创建了一个名为 LinkedList 的类：
class LinkedList
private Node head;
private int count;
public LinkedList()
this.head = null;
this.count = 0;
public bool Empty
get return this.count == 0;
public int Count
get return this.count;
public object this[int index]
get return this.Get(index);
public object Add(int index,object o)
if (index < 0)
throw new ArgumentOutOfRangeException("Index - " + index); //if index is less than 0 throw an error message
if (index > count) // if size exceeds the limit of the list the item will be added to the last line of the list.
index = count;
Node current = this.head;
if(this.Empty || index== 0)
this.head = new Node(o, this.head);
else
for(int i = 0; i < index - 1; i++)
current = current.Next;
current.Next = new Node(o, current.Next);
count++;
return o;
public object Add(Object o)
return this.Add(count, o);
public object Remove(int index)
if (index < 0)
throw new ArgumentOutOfRangeException("Index - " + index);
if (this.Empty)
return null;
if (index >= this.count)
index = count-1;
Node current = this.head;
object result = null;
if (index == 0)
result = current.Data; //gets the first node
this.head = current.Next; //makes 2nd node to the first node
else
for(int i = 0; i < index - 1; i++)
result = current.Next.Data;
result = current.Next;
current.Next = current.Next.Next;
count--;
return result;
public int IndexOf(object o)
Node current = this.head;
for(int i = 0; i < this.count; i++)
if (current.Data.Equals(o))
return i;
current = current.Next;
return -1;
public bool Contains(object o)
return this.IndexOf(o) >= 0; //if list contains object it returns bigger value than -1 and also 0.
public object Get(int index)
if(index < 0)
throw new ArgumentOutOfRangeException("Index - " + index);
if (this.Empty)
return null;
if(index >= this.count)
index = this.count-1;
Node current = this.head;
for(int i=0;i< index; i++)
current = current.Next;
return current.Data;
还有一个名为“Node”的类：
class Node
private object data;
private Node next;
public Node(object data,Node next) //constructor
this.data = data;
this.next = next;
public object Data
get return this.data;
set this.data = value;
public Node Next
get return this.next;
set this.next = value;
在主程序中，我从linkedlist类创建了两个对象，从用户那里读取了两个字符串，并将单词的字符添加到链表中。然后比较了这些字符，如果他们发现它们将从链表中删除，增加了计数器等等。如果计数器等于list1的元素数，那么它们是字谜，如果不是的话，则不是字谜。这是我的主程序代码：
class Program
static void Main(string[] args)
int counter = 0;
String word1, word2;
Console.WriteLine("Welcome to Anagram Checker!\nPlease enter your first word:");
word1 = Console.ReadLine();
Console.WriteLine("\nPlease enter the second word:");
word2 = Console.ReadLine();
int result = AnagramChecker(word1, word2, counter);
if (result == 1)
Console.WriteLine("These words are anagram");
if (result == 0)
Console.WriteLine("The words are not anagrams");
Console.ReadLine();
public static int AnagramChecker(String word1, String word2, int counter)
char[] ArrayWord1 = word1.ToCharArray();
char[] ArrayWord2 = word2.ToCharArray();
LinkedList list1 = new LinkedList();
LinkedList list2 = new LinkedList();
for (int i = 0; i < ArrayWord1.Length; i++) //Adds char of word1 to the list
list1.Add(i,ArrayWord1[i]);
for (int j = 0; j < ArrayWord2.Length; j++) //Adds char of word2 to the list
list2.Add(j,ArrayWord2[j]);
int max;
if (list1.Count >= list2.Count)
max = list1.Count;
if (list2.Count > list1.Count)
max = list2.Count;
for (int i = 0; i < list1.Count; i++)
if (list2.Contains(list1[i]) && list1.Contains(list2[i]))
list1.Remove(i);
list2.Remove(list2.IndexOf(list1[i]));
counter++;
Console.WriteLine(counter);
if (counter == word1.Length || counter == word2.Length)
return 1;
else
return 0;
当我输入不同的单词时，我得到不同的结果。输出示例如下。我做错了什么？
输出：
1-)
2-)
感谢您的帮助。
【问题讨论】：
您是否真的需要使用所有这些来确定 2 个单词是字谜还是您愿意接受更短的方法？ 当然，我对更短的方法持开放态度。我只是想测试一下自己，这就是我编写所有这些方法的原因。 你可以考虑这个问题：codereview.stackexchange.com/questions/127580/… 【参考方案1】：如果你只是想找出单词是否是字谜，你可以使用这个方法：
private static bool areAnagrams(string word1, string word2)
List<char> w1 = word1.OrderBy(c => c).ToList();
List<char> w2 = word2.OrderBy(c => c).ToList();
return !w1.Where((t, i) => t != w2[i]).Any();
这会创建两个按单词 chars 排序的列表，然后比较两者。
更具可读性的等效项：
private static bool areAnagrams(string word1, string word2)
List<char> w1 = word1.OrderBy(c => c).ToList();
List<char> w2 = word2.OrderBy(c => c).ToList();
if (w1.Count != w2.Count)
return false;
for (int i = 0; i < w1.Count; i++)
if (w1[i] != w2[i])
return false;
return true;
【讨论】：
【参考方案2】：我解决了这个问题，我只是修改了检查它们是否是字谜的 if 语句：
for (int i = 0; i < list1.Count; i++)
if (list2.Contains(list1[i]))
list1.Remove(i);
// list2.Remove(list2.IndexOf(list1[i]));
i--;
counter++;
感谢大家的帮助:)
【讨论】：
FIY:AnagramChecker("oooo", "toto", 0) 返回 True 但它不应该，而且你的程序中没有使用很多代码（max，result），你应该了解自动属性。祝你好运！【参考方案3】：
你的答案是：
int[] sayilar1 = new int[150];
int[] sayilar2 = new int[150];
public Form1()
InitializeComponent();
private void Form1_Load(object sender, EventArgs e)
var rand = new Random();
for (int i = 0; i < sayilar1.Length; i++)
sayilar1[i] = rand.Next();
sayilar2[i] = rand.Next();
lvNumbers.Items.Add(sayilar1[i].ToString());
lvNumbers.Items.Add(sayilar2[i].ToString());
private void btnShuffle_Click(object sender, EventArgs e)
int[]newArray=BubbleSort();
for (int i = 0; i < newArray.Count(); i++)
lvSorted.Items.Add(newArray[i].ToString());
private int[] BubbleSort()
int temp = 0;
int[] newArray = new int[300];
for (int i = 0; i < 300; i++)
if (i < 150)
newArray[i] = sayilar1[i];
if (i >= 150)
newArray[i] = sayilar2[i - 150];
for (int i = 0; i < newArray.Length; i++)
for (int sort = 0; sort < newArray.Length - 1; sort++)
if (newArray[sort] > newArray[sort + 1])
temp = newArray[sort + 1];
newArray[sort + 1] = newArray[sort];
newArray[sort] = temp;
return newArray;
2.
private void btnTek_Click(object sender, EventArgs e)
lvFiltered.Items.Clear();
string[] sayilar = tbSayilar.Text.Split('\n');
int[] array = new int[sayilar.Length];
for (int i = 0; i < sayilar.Length; i++)
array[i] = int.Parse(sayilar[i]);
List<int> ayiklanmisSayilar = TekCiftAyir(array, "T");
for (int i = 0; i < ayiklanmisSayilar.Count; i++)
lvFiltered.Items.Add(ayiklanmisSayilar[i].ToString());
private void btnCift_Click(object sender, EventArgs e)
lvFiltered.Items.Clear();
string[] sayilar = tbSayilar.Text.Split('\n');
int[] array = new int[sayilar.Length];
for (int i = 0; i < sayilar.Length; i++)
array[i] = int.Parse(sayilar[i]);
List<int> ayiklanmisSayilar = TekCiftAyir(array, "C");
for (int i = 0; i < ayiklanmisSayilar.Count; i++)
lvFiltered.Items.Add(ayiklanmisSayilar[i].ToString());
private List<int> TekCiftAyir(int[] array, string TC)
List<int> ayiklanmisSayilar = new List<int>();
if (TC == "T")
for (int i = 0; i < array.Length; i++)
if (array[i] % 2 == 1)
ayiklanmisSayilar.Add(array[i]);
if (TC == "C")
for (int i = 0; i < array.Length; i++)
if (array[i] % 2 == 0)
ayiklanmisSayilar.Add(array[i]);
return ayiklanmisSayilar;
【讨论】：
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