【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。
标签： c# linked-list anagram