C# Anagram Checker 与 LinkedList

Question

我正在尝试检查两个单词是否是变位词并尝试使用 LinkedList.To 来做到这一点，首先，我创建了一个名为 LinkedList:

的 class

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;
    }
}

还有另一个 class 名为 "Node":

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; }
    }
}

并且在主程序中，我从链表 class 创建了两个对象并从用户读取了两个字符串并将单词的字符添加到链接中 list.And 比较了这些字符，如果他们发现了它们'将从链表中删除，增加计数器，因此 on.If 计数器等于 list1 的元素数，如果不是的话，它们就是变位词我的主要程序代码：anagrams.Here

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;
    }
}

当我输入不同的词时，我得到不同的 results.The 输出示例是 below.What 我做错了吗？

输出：

1-)

2-)

感谢您的帮助。

Answer 1

如果您只是想知道单词是否是变位词，您可以使用这种方法：

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();
}

这会创建两个以单词 chars 排序的列表，然后比较两者。

更具可读性的等价物：

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;
}

Answer 2

我解决了这个问题，我只是修改了检查它们是否是变位词的 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:)

Answer 3

您的答案是：

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;
    }


    }

C# Anagram Checker 与 LinkedList

C# Anagram Checker With LinkedList

c#

linked-list

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