在我的代码中在哪里以及如何实现分隔符？

Question

我制作的程序能够读取文件并打印所有不同单词的列表。但是，我想使用分隔符删除单词开头和结尾的所有逗号和其他字符。但是，我只是不知道如何以正确的方式构建它以及将它放在我的代码中的什么位置。有人可以解释一下在哪里以及如何正确实现它吗？

package nl.ru.ai.SjoerdSam.exercise7;

import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.util.Scanner;

public class Concordances
{

  final static int MAX_NR_OF_WORDS=20000;
  public static void main(String[] args) throws FileNotFoundException
  {
    try
    {
      String[] words=new String[MAX_NR_OF_WORDS];
      int[] freqs=new int[MAX_NR_OF_WORDS];
      boolean terminate=true;
      Scanner scanner=openTextFile();
      int nr=findAndCountWords(scanner,words,freqs);
      while(terminate)
      {
        Scanner input=new Scanner(System.in);
        System.out.println("Please enter 'read' to start reading a file and display number of words read, "+"'content' to display content (all currently stored words in the order of apperance), "+"'stop' to stop the program or "+"'count'+ the word you want to count to count the number of occurences of a word, "+"the total number of words and the percantage of occurences.");
        String command=input.nextLine();

        switch(command)
        {
          case "read":
            System.out.println("The number of words in this file is"+nr);
            break;
          case "content":
            displayWords(nr,words,freqs);
            break;
          case "stop":
            terminate=false;
            System.out.println("Program terminated");
            break;
          case "count":
            // Bit stuck here on how to do the count and show the frequency of a single word. if i would actually get the frequency the percentage could be found by dividing the frequency with total number of words found above
            Scanner single=new Scanner(System.in);
            System.out.println("Please type in the word you want to know data of");
            String word=single.nextLine();
            findAndCountWord(scanner,words,word);
            System.out.println("The frequency for the word"+" "+single+" "+"is"+findAndCountWord(single,words,word));
            break;

        }
      }
    }
    catch(IllegalArgumentException e)
    {
      System.out.print(e);
    }
  }
  static Scanner openTextFile() throws FileNotFoundException
  {
    assert (true);
    Scanner input=new Scanner(System.in);
    System.out.println("Please enter file name: ");
    String fileName=input.nextLine();
    FileInputStream inputStream=new FileInputStream(fileName);
    return new Scanner(inputStream);
  }
  static int findAndCountWords(Scanner scanner, String[] words, int[] freqs)
  {
    assert words!=null&&freqs!=null;
    int nr=0;
    while(scanner.hasNext())
    {
      String word=scanner.next();
      if(updateWord(word,words,freqs,nr))
        nr++;
    }
    return nr;
  }
  static boolean updateWord(String word, String[] words, int[] freqs, int nr)
  {
    assert nr>=0&&words!=null&&freqs!=null;
    int pos=sequentialSearch(words,0,nr,word);
    if(pos<nr)
    {
      freqs[pos]++;
      return false;
    } else
    {
      words[pos]=word;
      freqs[pos]=1;
      return true;
    }
  }
  static int sequentialSearch(String[] array, int from, int to, String searchValue)
  {
    assert 0<=from&&0<=to : "Invalidbounds";
    assert array!=null : "Array shouldbeinitialized";
    if(from>to)
      return to;
    int position=from;
    while(position<to&&(!array[position].equals(searchValue)))
      position++;
    return position;
  }
  static void displayFrequencies(int nr, String[] words, int[] freqs)
  {
    assert nr>=0&&words!=null&&freqs!=null;

    for(int i=0;i<nr;i++)
    {
      System.out.println(words[i]+" "+freqs[i]);
    }
  }
  static void displayWords(int nr, String[] words, int[] freqs)
  {
    assert nr>=0&&words!=null&&freqs!=null;

    for(int i=0;i<nr;i++)
    {
      System.out.println(words[i]);
    }
  }

  static int findAndCountWord(Scanner scanner, String[] words, String word)
  {
    assert words!=null;
    int wordCount=0;
    while(scanner.hasNext())
    {
      for(int i=0;i<words.length;i++)
      {
        if(word.equals(words[i]))
        {
          wordCount++;
        }
      }
    }
    return wordCount;
  }
}

此时的输出示例：

U.S.
state's
laws.
principal
office
4557
Melan
Dr.
S.
Fairbanks,
AK,
99712.,
scattered
throughout
numerous
locations.
809
North
1500
West,
Salt
Lake
City,
UT
84116,
(801)
596-1887,
email
business@pglaf.org.
Email
contact
information
http://pglaf.org
information:
Gregory
B.
Newby
Chief
Executive
Director
gbnewby@pglaf.org
4.
Donations

Answer 1

这不是最漂亮的，但它应该可以工作。

String str = "!!!!@word's!@#";
String temp = "";
boolean done = false;
for (int i = 0; i < str.length(); i++) {
    char currentChar = str.charAt(i);
    if (Character.isLetter(currentChar) && !done) {
        done = true;
        temp += currentChar;
    } else if (done) {
        temp += currentChar;
    }
}

str = "";
done = false;
for (int i = temp.length() - 1; i >= 0; i--) {
    char currentChar = temp.charAt(i);
    if (Character.isLetter(currentChar) && !done) {
        str += currentChar;
        done = true;
    } else if (done) {
        str += currentChar;
    }
}
StringBuilder sb = new StringBuilder(str);

str = sb.reverse().toString();
System.out.println(str);

输出是单词的