我無法弄清楚如何讓遊戲重新啓動。我想我需要把它放在System.out.println(「好吧,讓我們再玩一次」)。 我需要在那裏重新啓動它?我錯過了什麼?一旦猜到數字,我該如何讓這個數字猜測遊戲重新啓動?
import java.util.Random;
import java.util.Scanner;
public class GuessMyNumber {
public static void main(String[] args) {
Random Bob = new Random();
int NumToGuess = Bob.nextInt(20);
int NumOfTries = 0;
Scanner Bill = new Scanner(System.in);
int guess;
boolean win = false;
while (win == false) {
System.out.println("I'm thinking of a number between 1 and 20, can you guess it?");
guess = Bill.nextInt();
NumOfTries++;
if (guess == NumToGuess) {
win = true;
}
else if (guess < NumToGuess) {
System.out.println("Your guess is too low");
}
else if (guess > NumToGuess) {
System.out.println("Your guess is too high");
}
}
System.out.println("Congratulations!");
System.out.println("The number was " + NumToGuess);
System.out.println("It took you " + NumOfTries + " tries");
while (win == true) {
String YesOrNo;
YesOrNo = Bill.nextLine();
System.out.println("Would you like to play again? " + YesOrNo);
YesOrNo = Bill.nextLine();
if (YesOrNo.equalsIgnoreCase("yes"))
{
System.out.println("Ok, let's play again.");
}
else
{
System.out.println("Ok, maybe next time.");
}
}
}
}
你缺少你在永恆的循環停留,而不是回到計劃流程的開始。 – EpicPandaForce 2014-12-07 21:44:02
@ crpittman18順便說一句,'''Bob.nextInt(20)''產生一個介於0(含)和20(不含)之間的數字。 – mezzodrinker 2014-12-07 21:51:33