java - 拼字游戏字谜生成器

Question

我正在尝试编写一个拼字游戏字谜生成器。

到目前为止，我的代码可以工作，但速度非常慢，并且存在错误。其中之一是它会多次使用字母。例如:输入字母:“ABCDEFG”。它会生成AB，但也会生成AA，这是不对的。

请帮忙。

public class Scrabble1
{
    private String[] dictionary2 = new String[97];
    private String[] dictionary3 = new String[978];
    private String[] dictionary4 = new String[3904];
    private String[] dictionary5 = new String[8635];
    private String[] dictionary6 = new String[15225];
    private String[] dictionary7 = new String[23097];
    public void sampleMethod(String s) throws FileNotFoundException
    {
        File in2 = new File( "dictionary2.txt" );
        File in3 = new File( "dictionary3.txt" );
        File in4 = new File( "dictionary4.txt" );
        File in5 = new File( "dictionary5.txt" );
        File in6 = new File( "dictionary6.txt" );
        File in7 = new File( "dictionary7.txt" );        
        Scanner dict2 = null,dict3 = null,dict4 = null,dict5 = null,dict6 = null,dict7 = null;

        try
        {
            dict2 = new Scanner(in2);
            dict3 = new Scanner(in3);   
            dict4 = new Scanner(in4);
            dict5 = new Scanner(in5);
            dict6 = new Scanner(in6);  
            dict7 = new Scanner(in7); 
            int c = 0;
            while(dict2.hasNext()&&dict3.hasNext()&&dict4.hasNext()&&dict5.hasNext()&&dict6.hasNext()&&dict7.hasNext())
            {
                dictionary2[c] = dict2.next();
                dictionary3[c] = dict3.next();
                dictionary4[c] = dict4.next();
                dictionary5[c] = dict5.next();
                dictionary6[c] = dict6.next();
                dictionary7[c] = dict7.next();
                c++;
            }
        }
        catch( FileNotFoundException e )
        {
            System.err.println( e.getMessage () );
            System.exit(1);
        }
        finally
        {
            dict2.close();
            dict3.close();
            dict4.close();
            dict5.close();
            dict6.close();
            dict7.close();
        }

       // for(int i= 0; i<80612; i++)
            //System.out.println(dicArray[i]);


        String temp = "";
        //All 2 letter anagrams  
        for(int k=0; k<=6; k++)
            for(int i=0; i<=6; i++)
                for(int d= 0; d<97; d++)
                {
                    temp = "" + s.charAt(k) + s.charAt(i);
                    if(temp.equals(dictionary2[d]))
                        System.out.println(temp  );
                }

        //All 3 letter anagrams  
        for(int j = 0; j<=6; j++)
            for(int k=0; k<=6; k++)
                for(int i=0; i<=6; i++)
                     for(int d= 0; d<978; d++)
                          {
                                temp = "" + s.charAt(j) + s.charAt(k)+ s.charAt(i);
                                if(temp.equals(dictionary3[d]))
                                    System.out.println(temp  );
                          }
        //All 4 letter anagrams  
        for(int j = 0; j<=6; j++)
            for(int k = 0; k<=6; k++)
                for(int i=0; i<=6; i++)
                    for(int l=0; l<=6; l++)
                          for(int d= 0; d<-3904; d++)
                          {
                                temp = "" + s.charAt(j) + s.charAt(k)+ s.charAt(i)+ s.charAt(l);
                                if(temp.equals(dictionary4[d]))
                                    System.out.println(temp );
                          }
         //All 5 letter anagrams
         for(int j = 0; j<=6; j++)
            for(int k = 0; k<=6; k++)
                for(int i=0; i<=6; i++)
                    for(int l=0; l<=6; l++)
                        for(int f=0; f<=6; f++)
                          for(int d= 0; d<8635; d++)
                          {
                                temp = "" + s.charAt(j) + s.charAt(k)+ s.charAt(i)+ s.charAt(l)+s.charAt(f);
                                if(temp.equals(dictionary5[d]))
                                    System.out.println(temp  );
                          }
          //All 6 letter anagrams
          for(int j = 0; j<=6; j++)
            for(int k = 0; k<=6; k++)
                for(int i=0; i<=6; i++)
                    for(int l=0; l<=6; l++)
                        for(int f=0; f<=6; f++)
                            for(int g=0; g<=6; g++)
                          for(int d= 0; d<15225; d++)
                          {
                                temp = "" + s.charAt(j) + s.charAt(k)+ s.charAt(i)+ s.charAt(l)+ s.charAt(f)+ s.charAt(g);
                                if(temp.equals(dictionary6[d]))
                                    System.out.println(temp  );
                          }
          //All 7 letter anagrams.
          for(int j = 0; j<=6; j++)
            for(int k = 0; k<=6; k++)
                for(int i=0; i<=6; i++)
                    for(int l=0; l<=6; l++)
                        for(int f=0; f<=6; f++)
                            for(int g=0; g<=6; g++)
                                for(int p=0; p<=6; p++)
                          for(int d= 0; d<23097; d++)
                          {
                                temp = "" + s.charAt(j) + s.charAt(k)+ s.charAt(i)+ s.charAt(l)+ s.charAt(f)+ s.charAt(g)+ s.charAt(p);
                                if(temp.equals(dictionary7[d]))
                                    System.out.println(temp  );

                          }




    }
}

字典文件仅按单词大小排序。

Answer 1

你可以构建一个trie从字典中取出并遍历它。对于输入字符串中的每个字符，转到 trie 中相应的节点，从输入中删除该字符并递归地重复。

伪代码:

function check(trie_node)
    if trie_node is terminal
        output trie_node
    else
        for each child of trie_node
            let c be the character of the child
            if input contains at least one c
                remove one c from input
                check(child)
                put c back into input
            end
        end
    end
end

check(trie_root)

您可以使用查找表来快速检查输入中还剩下多少个特定字符(恒定时间检查)。