「Play Again？」を追加するJavaでの私の乱数推測ゲームの機能

Question

私はこれに固執されており、ユーザーに続行を求められたときに再びループしているようには見えません。 do-whileループは、このタイプの再生機能に適していますか？どんな助けもありがとう、ありがとう！

import java.util.Scanner; 
import java.util.Random; 

public class HiLo 

{ 
    public static void main(String[] args) 
{ 
    boolean again = true; 

    while (again == true) 
    { 
    int MAX = 100; 
    int MIN = 1; 
    int guessCounter = 1; 
    int outOfRangeCounter = 0; 
    int userGuess = 0; 
    Random rand = new Random(); 
    int computerGuess = rand.nextInt(MAX) + MIN; 
    System.out.printf("Welcome to the game of Hi-Lo...\n\n\n"); 
    System.out.printf("I have chosen a random number for you to guess.\n\n"); 
    Scanner keyboard = new Scanner(System.in); 
    String repeatAgain; 


    while (userGuess != computerGuess) 
    { 
    System.out.printf("Guess %d: Enter a number between %d and %d: ",guessCounter,MIN,MAX); 
    userGuess = keyboard.nextInt(); 
    if (userGuess == computerGuess) 
    { 
     System.out.printf("%d is correct\n\n",userGuess); 
    } 
    else if (userGuess > MAX || userGuess < MIN) 
    { 
     System.out.printf("%d is not between %d and %d\n\n",userGuess,MIN,MAX); 
     outOfRangeCounter++; 
    } 
    else if (userGuess > computerGuess) 
    { 
     System.out.printf("%d is too high\n\n",userGuess); 
     MAX = userGuess - 1; 
     guessCounter++; 
    } 
    else if (userGuess < computerGuess) 
    { 
     System.out.printf("%d is too low\n\n",userGuess); 
     MIN = userGuess + 1; 
     guessCounter++; 
    }  
    } 
    System.out.printf("It took you %d valid guesses to find the number.\n",guessCounter); 
    System.out.printf("You had %d out of range guesses.\n\n",outOfRangeCounter); 
    System.out.printf("Do you want to play again? (Y or N): "); 
    repeatAgain = keyboard.nextLine(); 
    if (repeatAgain.equalsIgnoreCase("Y")) 
     { 
      again = true; 
     } 
    else 
     { 
      again = false; 
     } 
    }   
} 
} 

Answer 1

変更repeatAgain = keyboard.next();とrepeatAgain = keyboard.nextLine();、それが固定されなければなりません。

Answer 2

nextLine（）はスキップされ、スキップされた行が返されます。

 //change 
     //repeatAgain = keyboard.nextLine(); 
     //to 
     repeatAgain = keyboard.next();