java - OOPS(JAVA)中的类设计。

有什么方法可以从接口中删除totalPermutationCount（），因为该方法的唯一目的是从具体类中获取permutationCombination值，如果存在更多具体类或更多实例变量，则接口将变得团团而庞大。
除了构造函数以外，还有没有其他最佳方法可以将值分配给整个类所依赖的实例变量。


请参阅我的使用方式，请帮助我变得更好。

接口

interface Word {
  void performPermutation();
  int totalPermutationCount();
}

接口实现

class WordImpl implements Word{

//  "word" is to store the word from the user
private String word;
// "convert" is to convert the word into a char array to perform the logic
private char[] convert;
// "permutationCombination" is to find the total number of combination that is done
private int permutationCombination = 0;

// Constructor
public WordImpl(String wordCom){
    setWord(wordCom);
}

//getter setter of word instance variable
public void setWord(String wordCom){
    if(!wordCom.isEmpty() && wordCom.trim().length() > 0){
    word = wordCom;
    }else{
        word = "Default";
    }
    convertToChar();
}

// convert the given word to char array
private void convertToChar(){
    convert = new char[word.length()];
    for(int i = 0 ; i < word.length() ; i++){
    convert[i] = word.charAt(i);
}
}

public int totalPermutationCount(){
    return permutationCombination;
}

// -------------------------------- Below is the Temporary Logic Ignore it  ---------------------------------------------
private void performSwap(char[] list , int from, int to){
    char temp;
    temp = list[from];
    list[from] = list[to];
    list[to] = temp;
}

public void performPermutation(){
    char[] list = convert;
    Set<String> listData = new HashSet<>();
    System.out.println(convert);
    for (int i = 0 ; i < word.length() ; i++){
        for (int j = i + 1 ,  reverse = i - 1 ; j < word.length() || reverse >= 0  ; j++ , reverse--){
            if(j < word.length()){
            performSwap(list,i,j);
            System.out.println(convertToString(list));
            list = convert;
            permutationCombination++;
            }
            if(reverse >= 0 && i != 0){
                performSwap(list,i,reverse);
                 System.out.println(convertToString(list));
                 list = convert;
                 permutationCombination++;
            }
        }
    }
}
 // ----------------------------------------------------------------------------------------
private String convertToString(char[] list){
    String value = "";
    for(int i = 0 ; i < word.length() ; i++){
    value = value +  list[i];
     }
    return value;
}}

主班

 public class MyClass {
    public static void main(String args[]) {
    Word wordImplReference = new WordImpl("home");
    wordImplReference.performPermutation();
    System.out.println(wordImplReference.totalPermutationCount());
    }
   }

3.如何解决突变问题
char [] list = convert;

public void performPermutation(){
        char[] list = convert;
        Set<String> listData = new HashSet<>();
        System.out.println(convert);
        for (int i = 0 ; i < word.length() ; i++){
            for (int j = i + 1 ,  reverse = i - 1 ; j < word.length() || reverse >= 0  ; j++ , reverse--){
                if(j < word.length()){
                performSwap(list,i,j);
                System.out.println(convertToString(list));
                list = convert;
                permutationCombination++;
                }
                if(reverse >= 0 && i != 0){
                    performSwap(list,i,reverse);
                     System.out.println(convertToString(list));
                     list = convert;
                     permutationCombination++;
                }
            }
        }
    }

最佳答案

以下可以帮助您进行查询：


如果从接口中删除totalPermutationCount方法，则会失去从接口类获得的抽象好处。如果仍然需要这样做，则需要使用WordImpl wordImplReference = new WordImpl（“ home”）;在您的主要方法中。
在这种情况下，通过构造函数进行分配会更好。可以使用WordImpl类中的Set方法进行设置，但需要在Interface中添加setWord方法。