英语链的接龙

接龙规则：

大家经常玩成语接龙游戏，我们试一试英语的接龙吧：一个文本文件中有N 个不同的英语单词， 我们能否写一个程序，快速找出最长的能首尾相连的英语单词链，每个单词最多只能用一次。

最长的定义是：最多单词数量，和单词中字母的数量无关。

代码：

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package file; import java.io.*; import java.util.*; public class ceshi {     public static void main(String[] args)throws FileNotFoundException, IOException  {           File file=new File("D:/《飘》英文版.txt");         File file2=new File("D:/output3707.txt");         if(!file.exists())         {             System.out.println("文件不存在！");         }         else if(file.exists() && file.length() == 0) {             System.out.println("文件为空！");         } //判断是否文件为空         else         {             BufferedReader br=new BufferedReader(new FileReader(file));             BufferedWriter bw = new BufferedWriter(new FileWriter(file2)) ;               String s="";             int i,j;             ArrayList<ArrayList<String>> listall=new ArrayList();             String key;             String str;             while ((str = br.readLine()) != null) {                 s=s+str;             }             /* s=br.readLine();*/             String s2 = s.replaceAll("\\s+", " ");             System.out.println(s2);             //将文本文件中的所有信息变为字符串             String []s1=s2.split(" ");//以空格为分割划分每个单词             if(s1.length==1)                 System.out.println("单词数为1");             else             {                   for(j=0;j<s1.length;j++)                 {                       ArrayList<String> list=new ArrayList<String>();                     key=s1[j].substring(s1[j].length()-1);                     list.add(s1[j]);                     System.out.println("key:"+key);                     for(i=1;i<s1.length;i++)                     {                           if(key.equals(s1[i].substring(0, 1)))                         {                             list.add(s1[i]);                             key=s1[i].substring(s1[i].length()-1);                         }                         else                             continue;                     }                     listall.add(list);                 }                 int max=0;                 int count=0;                   for(i=0;i<listall.size();i++)                 {                     if(max<listall.get(i).size())                     {                         max=listall.get(i).size();                         count=i;                     }                 }                 for(i=0;i<listall.get(count).size();i++)                 {                     System.out.println(listall.get(count).get(i));                     bw.write(listall.get(count).get(i));                     bw.write(" ");                 }                 br.close();                 bw.close();             }         }     } }