數組大小聲明

Question

我已經在下面發佈了我的代碼。我在聲明數組wrongAnswers的行有一個問題。我以前能夠讓我的代碼工作，但問題是有些人自己刪除了我的所有文件。我能夠在不使用List或ArrayList的情況下工作。我只想了解，在嘗試使用其他方法之前，我現在可以如何開始工作。我明白Java數組是不可變的。不過，我仍然可以在以前的某個方面做到這一點。如果有人能幫我弄清楚我以前做過什麼，我會非常感激。

private Scanner keyboard = new Scanner(System.in); 

private final String[] testAnswers = { 
     "B","D","A","A","C", 
     "A","B","A","C","D", 
     "B","C","D","A","D", 
     "C","C","B","D","A"}; 
private String[] studentAnswers = new String[20]; 
/* 
private String[] studentAnswers = { 
     "B","D","A","A","C", 
     "A","B","A","C","D", 
     "B","C","D","A","D", 
     "C","C","B","D","A"}; 
*/ 
private int[] wrongAnswers; 
private int answeredCorrectly = 0; 

public void getStudentAnswers() { 
    for(int x = 0; x < 20; x++) { 
     do { 
      System.out.print("Enter answer for #" + (x + 1) + " : "); 
      this.studentAnswers[x] = keyboard.next().toUpperCase(); 
      if (!"A".equals(this.studentAnswers[x]) && 
        !"B".equals(this.studentAnswers[x]) && 
        !"C".equals(this.studentAnswers[x]) && 
        !"D".equals(this.studentAnswers[x])) { 
       System.out.println("Invalid input."); 
      } 
     } while(!"A".equals(this.studentAnswers[x]) && 
       !"B".equals(this.studentAnswers[x]) && 
       !"C".equals(this.studentAnswers[x]) && 
       !"D".equals(this.studentAnswers[x])); 
    } 
} 

public int totalCorrect() { 
    int arrayLocation = 0; 

    for(int x = 0; x < 20; x++) { 
     if (this.studentAnswers[x].equals(this.testAnswers[x])) 
      this.answeredCorrectly++; 
     else 
      this.wrongAnswers[arrayLocation++] = x; 
    } 

    return this.answeredCorrectly; 
} 

public int totalIncorrect() { 
    return 20 - this.answeredCorrectly; 
} 

public boolean passed() { 
    return this.answeredCorrectly >= 15; 
} 

public void questionsMissed() { 
    if(this.answeredCorrectly != 20) { 
      for(int x = 0; x < this.wrongAnswers.length; x++) { 
      System.out.println(this.wrongAnswers[x]); 
     } 
    } 
} 

Answer 1

如果代碼編寫得很好，節省空間（這是您要做的）通常會降低性能，反之亦然。正如你所看到的，你可以達到你想要的，但是你會失去表現。

我發現演繹在解決類似問題時很有用。條件：

1）數組是不可變的 2）要分配的，你需要

2點提出了一個問題空間的確切數額：你怎麼知道你需要多大的空間？明顯的答案：知道你有多少（正確）答案。從那裏你可以做：

public int totalCorrect() { 
    for(int x = 0; x < 20; x++) { 
     if (this.studentAnswers[x].equals(this.testAnswers[x])) 
      this.answeredCorrectly++; 
    } 

    this.wrongAnswers = int[20 - this.answeredCorrectly]; 

    // Here you want to create your wrongAnswers, but you have to go over 
    // the same array twice... 
    int arrayLocation = 0; 
    for(int x = 0; x < 20; x++) { 
     if (!this.studentAnswers[x].equals(this.testAnswers[x])) 
      this.wrongAnswers[arrayLocation++] = x; 
    } 


    return this.answeredCorrectly; 
} 

有可能有更多的方法來做類似的事情，並取得更好的表現。乍一看，他們看起來像我不喜歡的方法，我會使用列表，如已經提出的，或者已經提出，或者一個集合，但誰知道...

Answer 2

private int [] wrongAnswers = new int [20] ;