如何更改我的代碼以提高效率？

Question

在這段代碼中是在計算機和用戶之間玩石頭剪刀布。 我的代碼運行良好，但是，我正在嘗試一種更好的方法來詢問用戶是否想再次玩。 如果是，那么它將再次啟動程序，如果不是，那么它將停止。 我的“是”似乎有效，但沒有將停止，而不是 go 一直到。 有關如何執行此操作的任何建議或提示？ 我將嘗試合並一個不同的 while 循環，但沒有工作。 do循環對這種情況有好處嗎？ 謝謝！

//import scanner
import java.util.Scanner;
import java.util.*;

//declare variables and main methods
class Rock {
  Scanner scan = new Scanner(System.in);
  Random generator = new Random();
  String response, name;
  char choice;
  int rounds, computerChoice, userScore, computerScore;
  boolean playIntro = true;
  boolean playGame = true;

  //this method will run the entire progrma
  public void playRPS(){
  
    //while loop for beginning of game
    while(playIntro){
      System.out.println("This is a game of Rock Paper Scissors!");
      System.out.println("Please enter your name: ");
      name = scan.nextLine();

      //while loop for the actual part of the game
      while(playGame){
        System.out.println("Type R (Rock), P (Paper), or S (Scissors): ");
        choice = scan.nextLine().charAt(0);
        computerChoice = generator.nextInt(3)+1;

        //using switch and case for each choice
        switch (choice){
          //case for Rock
          case 'R':
            if(computerChoice==1){
             System.out.println("Tie between you and the computer! Go again.");
             break;
            }
            else{
              if(computerChoice==2){
                System.out.println("The computer beat you this round");
                computerScore++;
                break;
              }
              else{
                System.out.println("You won this round");
                userScore++;
                break;
              }
            }
          //case for Paper
          case 'P':
            if(computerChoice==2){
             System.out.println("Tie between you and the computer! Go again.");
             break;
            }
            else{
              if(computerChoice==3){
                System.out.println("The computer beat you this round");
                computerScore++;
                break;
              }
              else{
                System.out.println("You won this round");
                userScore++;
                break;
              }
            }
          //case for Scissors
          case 'S':
            if(computerChoice==3){
             System.out.println("Tie between you and the computer! Go again.");
             break;
            }
            else{
              if(computerChoice==1){
                System.out.println("The computer beat you this round");
                computerScore++;
                break;
              }
              else{
                System.out.println("You won this round");
                userScore++;
                break;
              }
            }
        }
        System.out.println("You have "+userScore+" points and the computer has "+computerScore+" points");
        if (userScore==5){
          System.out.println("\nOut of 5 rounds, You beat the computer!");
          playGame = false;
        }
        else if (computerScore==5){
          System.out.println("\nOut of 5 rounds, The computer beat you.");
          playGame = false;
        }
      }
      askUser();
    }  
  }

  public void askUser(){

    System.out.println("\nDo you want to play this Rock Paper Scissors again? Type yes: ");
    response = scan.nextLine();
    if (response.equalsIgnoreCase("yes")){
      playGame = true;
      userScore=0;
      computerScore=0;
    }
    else{
      playGame = false;
      scan.nextLine();
    }
  }

  public static void main() {
    Rock prog = new Rock();
    prog.playRPS();
  }
}

Answer 1

我不會說這必然更有效甚至更好，但它更簡潔一些。 它的主要元素是。

• 使用 Lambdas 根據所選擇的動作來決定獲勝者。
• 根據用戶的移動，使用 map 調用正確的 Lambda。 lambda 然后評估這兩個動作來決定結果。
• 為簡單起見，移動是按數字選擇的

當然，重要的是您的代碼可以正常工作。

import java.util.List;
import java.util.Map;
import java.util.Random;
import java.util.Scanner;
import java.util.Set;
import java.util.function.Function;

public class RockPaperScissors {
    
    final static int PAPER = 1;
    final static int ROCK = 2;
    final static int SCISSORS = 3;
    
    // the next two declarations allows the previous three to take on any values.
    private Set<Integer> allowedMoves = Set.of(PAPER, ROCK, SCISSORS);
    private List<String> moves = List.of("PAPER", "ROCK", "SCISSORS");
    private String ROCK_WINS_MSG = "Rock crushes scissors";
    private String SCISSORS_WINS_MSG = "Scissors cuts paper";
    private String PAPER_WINS_MSG = "Paper covers rock";
    private String COMPUTER_WINS = ", computer wins!";
    private String YOU_WIN = ", you win!";
    
    private Function<Integer, String> CHECK_PAPER =
            (c) -> c == PAPER ? "It's a tie!" :
                    c == ROCK ? PAPER_WINS_MSG + YOU_WIN :
                    SCISSORS_WINS_MSG + COMPUTER_WINS;
    private Function<Integer, String> CHECK_ROCK =
            (c) -> c == ROCK ? "It's a tie!" :
                    c == SCISSORS ? ROCK_WINS_MSG + YOU_WIN :
                    PAPER_WINS_MSG + COMPUTER_WINS;
    private Function<Integer, String> CHECK_SCISSORS =
            (c) -> c == SCISSORS ? "It's a tie!" :
                    c == PAPER ? SCISSORS_WINS_MSG + YOU_WIN :
                    ROCK_WINS_MSG + COMPUTER_WINS;
    
    private Map<Integer, Function<Integer, String>> evalUser =
            Map.of(PAPER, CHECK_PAPER, ROCK, CHECK_ROCK, SCISSORS,
                    CHECK_SCISSORS);
    
    public static void main(String[] args) {
        new RockPaperScissors().play();
    }
    
    public void play() {
        Random r = new Random();
        Scanner scan = new Scanner(System.in);
        
        while (true) {
            System.out.printf("%n%d : %s%n%d : %s%n%d : %s%n%s%n",
                    PAPER, "PAPER", ROCK, "ROCK", SCISSORS,
                    "SCISSORS", "Any other integer to quit.");
            System.out.print("Your move! ");
            String str = scan.nextLine();
            int move;
            try {
                move = Integer.parseInt(str);
                if (!allowedMoves.contains(move)) {
                    break;
                }
            } catch (IllegalArgumentException ie) {
                System.out.println("Only integers permitted.");
                continue;
            }
            System.out.println("\nYou chose " + moves.get(move - 1));
            int cmove = r.nextInt(3);
            
            System.out.println(
                    "The computer chooses " + moves.get(cmove));
            System.out.println(evalUser.get(move).apply(cmove + 1));
            
        }
        System.out.println("\nGame over!");
        
    }
}

一旦建議您的代碼將是查看開關案例。 每種情況的代碼實際上是相同的。 我會尋找相似之處並將評估作為一種方法（我在代碼中並沒有真正做到這一點）。 然后在每種情況下，使用適當的 arguments 調用該方法。 一個這樣的論點將是基於上下文的“計算機”或“你”。

Answer 2

沒有任何東西可以設置playIntro假，因此外循環永遠不會終止。

當askUser()將playGame設置為 false 時，內部循環終止，並且您進入外部循環，該外部循環繼續循環。

我根本看不出外循環存在的任何理由。 您只想打印介紹並詢問玩家的姓名一次。

這與其說是“效率”問題，不如說是正確性問題。

---

順便說一句，最好讓askUser()返回一個真/假值，而不是設置一個成員變量。 然后你可以直接在'while'表達式中使用它。

playRPS()的整體結構如下所示：

public void playRPS() {
   ... print intro, ask name ...
   do {
       ... play one game ...
   } while (askUser());
}