Java 程序。 需要簡化 public boolean equals (object B){ is to long。 我不知道如何減少程序那部分的編碼

Question

我想減少公共 boolean 等於（對象 B）中的編碼。 程序正在運行。 output 沒問題。 同時創建一個新的文本文件（“outpuPalindrome”）。 我整個早上都在這個程序中工作，並且可以按照我的說明找到簡化 boolean 等於部分的編碼的方法。 如果有人能給出一個想法，我將不勝感激。

package charstack;

//import java.io.*; 
//import java.util.*; 
import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.Stack;

/*  StringToStack function reads the string into its private stack.
        Precondition: check if a character is an alphabetic letter
        Postcondition: only alphabetic letters of the string are in the private stack.
        Other characters(like space, digit, punct, etc) can not be in the stack.
*/
public class CharStack {
    private Stack<Character> myCharacters; 
    
    public CharStack(){
        myCharacters = new Stack<>();
    }
    public void StringToStack(String inStr){
        myCharacters = new Stack<>();
        for (Character c: inStr.toCharArray()){
            if (Character.isAlphabetic(c)){
                myCharacters.push(c);
            }
        }
    }
    
    /*  Reverse function empties its private stack out into another stack, 
        causing the order of the characters to be reversed.  Returns this new stack.
        Precondition:  Private stack is not empty.
    Postcondition:  If private stack is empty, displays error and returns.
        Otherwise: Returns new stack containing private stack's
           elements in reverse order.  Private stack is empty.
    */  
    public CharStack Reverse(){
        
        if (myCharacters.empty()){
            System.out.println("Error: CharStack is empty.");
            return null;
        }
        CharStack s = new CharStack();
        while (!myCharacters.empty()){
            s.myCharacters.push(myCharacters.pop());
        }
        return s; //return new CharStack();
    }
    
    
    
    /* IsEqual function tests if both private stack and B's stack are the same.
        Precondition:  Neither stack is empty.
    Postcondition:  If either stack is empty,
                displays error message and returns.
        Otherwise:
            Returns true if both stacks are the same, returns false otherwise.
            Both the private stack and B's stack are empty.
    */  
    @Override
    public boolean equals (Object B){
        
        if (myCharacters.empty()){
            System.out.println("Error: CharStack 1 is empty.");
            return false;
        }
        if (this == B){
            myCharacters = new Stack<>();
            return true; //return true;
        }
        if (B instanceof CharStack){
            CharStack bStack = (CharStack) B;
            if (bStack.myCharacters.empty()){
                System.out.println("Error: CharStack 2 is empty.");
                myCharacters = new Stack<>();
                return false;
            }
            while (!myCharacters.empty()){
                Character cA = myCharacters.pop();
                if (bStack.myCharacters.empty()){
                    myCharacters = new Stack<>();
                    bStack.myCharacters = new Stack<>();
                    return false;
                }
                Character cB = bStack.myCharacters.pop();
                if (cA != cB){
                    myCharacters = new Stack<>();
                    bStack.myCharacters = new Stack<>();
                    return false;
                }
            }
            if (!bStack.myCharacters.empty()){
                bStack.myCharacters = new Stack<>();
                return false;
            }
            return true;
        }
        myCharacters = new Stack<>();
        return false;
    }
    
    /* Postcondition:  Contents of stack have been converted to string on one line.  
       the private stack myCharacters is unchanged.
    */
    @Override
    public String toString(){
        //return "hello";
        StringBuilder sb = new StringBuilder();
        Stack<Character> temp = new Stack<>();
        while (!myCharacters.empty()){
            temp.push(myCharacters.pop());
        }
        while (!temp.empty()){
            Character c = temp.pop();
            myCharacters.push(c);
            sb.append(c);
        }
        return sb.toString();
    }
}

測試驅動文件.java

/*prompt user to enter file name for palindrome check
//(use provided palindrome1.txt and palindrome2.txt files)
//read file line by line and use the CharStack class to check 
//if the line is palindrome, you need to call StringToStack() function to 
//push all letters from the line to a CharStack
//output the stack using toString() function
//then call reverse() function to get the reverse of the above CharStack
//call equals() function to check if the contents of two CharStacks 
//are the same and make decision about palindrome
//output all palindrome lines to a file to save them*/

公共 class CharStackApp {

 public static void main(String[] args) throws FileNotFoundException {
    Scanner input = new Scanner(System.in);
    System.out.println("Please, enter Palindrome file text to read.");
    String str = input.nextLine();
    File palindrome = new File(str);
    File outputPalindrome = new File("OutputPalindrome");
    PrintWriter writer;
    Scanner fileScanner;
    try{
        fileScanner = new Scanner(palindrome);
        writer = new PrintWriter(outputPalindrome);
    }
    catch (FileNotFoundException e){
        return;
    }
    while (fileScanner.hasNextLine()){
        String file = fileScanner.nextLine().toLowerCase();
        CharStack s = new CharStack();
        
        s.StringToStack(file);// original (ln) yo puse file
        System.out.println(s);
        
        CharStack sRev = new CharStack();
        sRev.StringToStack(file);
        sRev = sRev.Reverse();
        
        if (s.equals(sRev)){
            writer.write(file);
            writer.write("\n");
        }
        writer.close();
    
    }

Answer 1

是的，這equals()實現方式有點矯枉過正。 只需比較堆棧：

@Override
public boolean equals(Object B) {
    if (! (B instanceof CharStack))
        return false;
    CharStack bStack = (CharStack) B;
    return this.myCharacters.equals(bStack.myCharacters);
}

請記住在實現hashCode() equals() 。 如果你養成這樣的習慣，以后會省去很多麻煩。

@Override
public int hashCode() {
    return this.myCharacters.hashCode();
}

您的toString()似乎也過於復雜。

@Override
public String toString() {
    StringBuilder sb = new StringBuilder();
    for (Character c : this.myCharacters)
        sb.append(c);
    return sb.toString();
}