使用哈希/列表輸出每個單詞句子

Question

我有一個程序，我有一個輸入的句子。該句子使用regEx/split方法分解成單詞。單詞存儲在名爲wordsInLine的數組中，我的工作是將這些單詞添加到名爲WordList的ArrayList中。

如果我們找到WordList和wordsInLine之間的匹配，我們繼續閱讀下一個單詞等等，等等。我的工作是輸出每個單詞。

預期輸入=遊戲的得分是五到五。

預期輸出：

一句話：

字：分數

字：中

一句話：

於是就等

主類：

個

package proj1; 

import java.util.ArrayList; 

public class Proj1 { 

    public static void main(String[] args) { 

    int lineWord = 0, listWord = 0; 

    String inputLine = "The score of the game is five to five"; 
    String regEx = "(, *)|(: *)|\\s"; 
    String[] wordsInLine; 

    ArrayList<Word> wordList = new ArrayList<Word>(); 

    wordsInLine = inputLine.split(regEx); 

    if (wordList.isEmpty()) 
     System.out.println("Empty List"); 



    } 
} 

詞類：

package proj1; 

class Word { 

    String Word; 
    int timesWordIsRepeated; 

    public Word (String words, int count) { 
    words = Word;  
    } 

    public String getWord() { 
    return Word; 
    } 

    @Override 
    public String toString() { 
    String theString = String.format("Word :",Word); 
    return theString; 
    } 
} 

Answer 1

可以遍歷這樣獲得的每個字：

String inputLine = "The score of the game is five to five"; 
for (String word : inputLine.split(" ")) { 
    System.out.println("Word: " + word); 
} 

打印：

Word: The 
Word: score 
Word: of 
Word: the 
Word: game 
Word: is 
Word: five 
Word: to 
Word: five 

Answer 2

HashMap可以更好地幫助您計算字那麼ArrayList

public static void main(String[] a) throws Exception { 

    String inputLine = "The score of the game is five to five"; 
    String regEx = "(, *)|(: *)|(\\? *)|(; *)|(! *)|(*\\(*)|(\\) *)|\\s"; 
    String[] wordsInLine; 

    Map<String, Integer> wordMap = new HashMap<String, Integer>(); 

    wordsInLine = inputLine.split(regEx); 

    for (String lineWord : wordsInLine) { 
     Integer count = wordMap.get(lineWord); 
     if (count == null) { 
      wordMap.put(lineWord, 1); 
     } else { 
      wordMap.put(lineWord, (count + 1)); 
     } 
    } 

    if (wordMap.isEmpty()) { 
     System.out.println("No entry found"); 
    } else { 
     for (Entry<String, Integer> entry : wordMap.entrySet()) { 
      // you can create Word pojo here if you want new Word(entry.getKey(), entry.getValue()) and add to some list 
      System.out.println(entry.getKey() + " : " + entry.getValue()); 
     } 
    } 

} 

UPDATE：

隨着ArrayList

public static void main(String[] a) throws Exception { 

     String inputLine = "The score of the game is five to five"; 
     String regEx = "(, *)|(: *)|(\\? *)|(; *)|(! *)|(*\\(*)|(\\) *)|\\s"; 
     String[] wordsInLine; 

     ArrayList<Word> wordList = new ArrayList<Word>(); 

     wordsInLine = inputLine.split(regEx); 

     for (String lineWord : wordsInLine) { 
      boolean wordFound = false; 
      for (Word word : wordList) { 
       if (word.getWord().equals(lineWord)) { 
        wordFound = true; 
        word.setCount(word.getCount() + 1); 
       } 
      } 
      if (!wordFound) { 
       wordList.add(new Word(lineWord, 1)); 
      } 
     } 
     if (wordList.isEmpty()) { 
      System.out.println("Empty List"); 
     } else { 
      for (Word word : wordList) { 
       System.out.println(word); 
      } 
     } 

    } 

    static class Word { 

     String word; 
     int count; 

     public Word(String word, int count) { 
      this.word = word; 
      this.count = count; 
     } 

     public String getWord() { 
      return word; 
     } 

     public void setWord(String word) { 
      this.word = word; 
     } 

     public int getCount() { 
      return count; 
     } 

     public void setCount(int count) { 
      this.count = count; 
     } 

     @Override 
     public String toString() { 
      return "Word [word=" + word + ", count=" + count + "]"; 
     } 

    } 

Answer 3

這是我會怎麼解決你的問題：

Map<String, Integer> wordCounts = new LinkedHashMap<>(); 

String inputLine = "The score of the game is five to five"; 
for (String word : inputLine.split(" ")) { 
    int count = 1; 
    if (wordCounts.containsKey(word)) { 
     count += wordCounts.get(word); 
    } 
    wordCounts.put(word, count); 
} 

for (Map.Entry<String, Integer> entry: wordCounts.entrySet()) { 
    System.out.println("Word: " + entry.getKey() + " (" + entry.getValue() + ")"); 
} 

打印：

Word: The (1) 
Word: score (1) 
Word: of (1) 
Word: the (1) 
Word: game (1) 
Word: is (1) 
Word: five (2) 
Word: to (1) 

要使用ArrayList你可以這樣做：

List<Word> words = new ArrayList<Word>(); 

String inputLine = "The score of the game is five to five"; 

outerLoop: 
for (String s : inputLine.split(" ")) { 
    for (Word existingWord : words) { 
     if (existingWord.value.equals(s)) { 
      existingWord.count++; 
      continue outerLoop; 
     } 
    } 
    words.add(new Word(s)); 
} 

for (Word word : words) { 
    System.out.println("Word: " + word.value + " (" + word.count + ")"); 
} 

有了這個類太：

class Word { 
    final String value; 
    int count = 1; 

    public Word(String value) { 
     this.value = value; 
    } 
} 

產地：

Word: The (1) 
Word: score (1) 
Word: of (1) 
Word: the (1) 
Word: game (1) 
Word: is (1) 
Word: five (2) 
Word: to (1)