如果元素包含空格，創建更長的數組？ （JAVA）

Question

目前，我想要做，做以下的方法：

• 需要3個字符串數組（也就是說，beforeList和afterList）
• 將查找在這兩個詞和詞beforeList，如果發現，與單詞替換afterList
• 返回一個新的數組，它本身

例如接通與afterList字符到新的元素的元素，這裏是一個測試的情況下，請注意，「我'm'分裂成兩個元素在最後一個陣列「我」和「上午」的最後一個排列：

String [] someWords = {"i'm", "cant", "recollect"}; 
    String [] beforeList = {"dont", "cant", "wont", "recollect", "i'm"}; 
    String [] afterList = {"don't", "can't", "won't", "remember", "i am"}; 
    String [] result = Eliza.replacePairs(someWords, beforeList, afterList); 
    if (result != null && result[0].equals("i") && result[1].equals("am") 
      && result[2].equals("can't") && result[3].equals("remember")) { 
     System.out.println("testReplacePairs 1 passed."); 
    } else { 
     System.out.println("testReplacePairs 1 failed."); 
    } 

我最大的問題是在佔這種空白的情況。我知道我將在下面發佈的代碼是錯誤的，但是我一直在嘗試不同的方法。我認爲我現在的代碼應該返回一個空數組，它是第一個數組的長度，但佔空間。我意識到它可能需要一個完全不同的方法。任何建議，雖然將不勝感激，我會繼續嘗試並弄清楚，但如果有辦法做到這一點，那麼我很樂意聽到並從中吸取教訓！謝謝。

public static String[] replacePairs(String []words, String [] beforeList, String [] afterList) { 
    if(words == null || beforeList == null || afterList == null){ 
     return null; 
    } 
    String[] returnArray; 
    int countofSpaces = 0; 
    /* Check if words in words array can be found in beforeList, here I use  
     a method I created "inList". If a word is found the index of it in      
     beforeList will be returned, if a word is not found, -1 is returned. 
     If a word is found, I set the word in words to the afterList value */ 
    for(int i = 0; i < words.length; i++){ 
     int listCheck = inList(words[i], beforeList); 
     if(listCheck != -1){ 
      words[i] = afterList[listCheck]; 
     } 
    } 
    // This is where I check for spaces (or attempt to) 
    for(int j = 0; j < words.length; j++){ 
     if(words[j].contains(" ")){ 
      countofSpaces++; 
     } 
    } 
    // Here I return an array that is the length of words + the space count) 
    returnArray = new String[words.length + countofSpaces]; 
    return returnArray; 

} 

Answer 1

下面是做它的衆多方法之一，假設你要處理這種情況的話包含連續超過1位情況：形式

for(int i = 0; i < words.length; i++){ 
     int listCheck = inList(words[i], beforeList); 
     if(listCheck != -1){ 
      words[i] = afterList[listCheck]; 
     } 
    } 

    ArrayList<String> newWords = new ArrayList<String>(); 

    for(int i = 0 ; i < words.length ; i++) { 
     String str = words[i]; 
     if(str.contains(' ')){ 
      while(str.contains(" ")) { 
       str = str.replace(" ", " "); 
      } 

      String[] subWord = str.split(" "); 

      newWords.addAll(Arrays.asList(subWord)); 
     } else { 
      newWords.add(str); 
     } 
    } 

    return (String[])newWords.toArray();