我的tic tac toe游戏在这里编写了我的java任务，一切都很好，除了一个小问题，当你进入最后一步（第九回合）时，最后一个'X'没有出现。这不仅令人讨厌，因为没有显示获胜的部分，但它导致了一些问题，领带方法没有得到正确解决，因此它没有显示任何内容。
我明确知道为什么不这样做，只需要帮助修复它。
do
{
ComputerSpotChoice = (int) (Math.random() * 9 ) + 1 ;
}
while (Board [(ComputerSpotChoice - 1) / 3] [(ComputerSpotChoice - 1) % 3] != '-') ;
当它进入第九回合时，这个条件总是为真，因为它找不到' - '。继续无限循环，我不太清楚如何解决这个问题，但保持计算机点选择半。
这是完整的代码。
/*I have methods for drawing the board, determining a winner, and a loser. This is just the 'main' method containing the bulk of the program */
public static void main (String[] args)
{
//Variable declaration
Scanner kbReader = new Scanner(System.in);
char [] [] Board = new char [3] [3] ;
String MenuInput ;
int BoardOutput ;
int UserSpotChoice ;
int ComputerSpotChoice = 0;
int UserTurn = 1 ;
int Winner = 0 ;
Board [0] [0] = '-' ;
Board [0] [1] = '-' ;
Board [0] [2] = '-' ;
Board [1] [0] = '-' ;
Board [1] [1] = '-' ;
Board [1] [2] = '-' ;
Board [2] [0] = '-' ;
Board [2] [1] = '-' ;
Board [2] [2] = '-' ;
//Welcome
System.out.println ("Welcome to Alex Montague's Tic Tac Toe game!") ;
System.out.println ("") ;
System.out.println ("If you wish to play, type 'Play'") ;
System.out.println ("If you wish to read the instructions, type 'Instructions'") ;
System.out.println ("If you wish to exit, type 'Exit'") ;
MenuInput = kbReader.next () ;
do
{
if (MenuInput.equals ("Play") || MenuInput.equals ("play"))
{
while (!GameOver)
{
System.out.println ("\f") ;
System.out.println (" Tic Tac Toe") ;
BoardOutput = DrawBoard (Board) ;
System.out.println (" 1 2 3") ;
System.out.println (" 4 5 6") ;
System.out.println (" 7 8 9") ;
System.out.println ("Please enter the number you would like to move your spot to") ;
UserSpotChoice = kbReader.nextInt () ;
if (UserSpotChoice == 1) Board [0] [0] = 'X' ;
if (UserSpotChoice == 2) Board [0] [1] = 'X' ;
if (UserSpotChoice == 3) Board [0] [2] = 'X' ;
if (UserSpotChoice == 4) Board [1] [0] = 'X' ;
if (UserSpotChoice == 5) Board [1] [1] = 'X' ;
if (UserSpotChoice == 6) Board [1] [2] = 'X' ;
if (UserSpotChoice == 7) Board [2] [0] = 'X' ;
if (UserSpotChoice == 8) Board [2] [1] = 'X' ;
if (UserSpotChoice == 9) Board [2] [2] = 'X' ;
do
{
ComputerSpotChoice = (int) (Math.random() * 9 ) + 1 ;
}
while
(Board [(ComputerSpotChoice - 1) / 3] [(ComputerSpotChoice - 1) % 3] != '-') ;
if (ComputerSpotChoice == 1) Board [0] [0] = 'O' ;
if (ComputerSpotChoice == 2) Board [0] [1] = 'O' ;
if (ComputerSpotChoice == 3) Board [0] [2] = 'O' ;
if (ComputerSpotChoice == 4) Board [1] [0] = 'O' ;
if (ComputerSpotChoice == 5) Board [1] [1] = 'O' ;
if (ComputerSpotChoice == 6) Board [1] [2] = 'O' ;
if (ComputerSpotChoice == 7) Board [2] [0] = 'O' ;
if (ComputerSpotChoice == 8) Board [2] [1] = 'O' ;
if (ComputerSpotChoice == 9) Board [2] [2] = 'O' ;
Winner (Board) ;
Loser (Board) ;
Tie (Board) ;
} //While loop
if (GameOver) System.exit (0) ;
} //If play
else if (MenuInput.equals ("Instructions") || MenuInput.equals ("instructions"))
{
System.out.println ("\f") ;
System.out.println ("You will be playing the game of Tic Tac Toe against the computer.") ;
System.out.println ("The object of this game is to get three of your own x's or o's in a line.") ;
System.out.println ("You take turns placing the x's and o's and whoever gets three in a row first wins.") ;
System.out.println ("Good Luck!") ;
System.out.println ("") ;
System.out.println ("If you wish to play, type 'Play'") ;
System.out.println ("If you wish to exit, type 'Exit'") ;
MenuInput = kbReader.next () ;
}
else if (MenuInput.equals ("Exit") || MenuInput.equals ("exit"))
{
System.out.println ("Thank you for using Alex Montague's Tic Tac Toe game!") ;
System.exit (0) ;
}
else
{
System.out.println ("Sorry, that is not a valid choice.") ;
System.out.println ("If you wish to play, type 'Play'") ;
System.out.println ("If you wish to read the instructions, type 'Instructions'") ;
System.out.println ("If you wish to exit, type 'Exit'") ;
MenuInput = kbReader.next () ;
}
} //do while
while (!MenuInput.equals ("Instructions") || !MenuInput.equals ("instructions") || !MenuInput.equals ("Play") || !MenuInput.equals ("play") || !MenuInput.equals ("Exit") || !MenuInput.equals ("exit")) ;
} // main method
答案 0 :(得分：0)
为什么不在开始时随机化一系列可能的动作，只需查看列表？那样你的循环必须结束。当你的当前循环不仅不能识别游戏结束时间，而且可以反复尝试相同的非法移动。
这是一种不涉及排序的方法：添加这些定义：
char free[9] = {1,2,3,4,5,6,7,8,9}; // choices not taken yet
int numFree = 9; // number of choices available
int moveIndex;
然后选择这样的举动：
moveIndex = (int) (Math.random() * numFree); // pick index of new move from free ones
ComputerSpotChoice = free[moveIndex]; // save that choice
numFree--; // there is now one less free move
free[moveIndex] = free[numFree]; // move the "last" free move to where you just took a move from
您需要搜索“免费”以供用户选择，以便将其删除。
答案 1 :(得分：0)
您的计划依靠机会找到最后一次移动。 这有点奇怪 - 你是否会在将数字随机拉出后将数字放回帽子中？ 不，你会让他们脱离帽子。
斯科特·亨特建议您首先列出随机排序的动作。继续这个比喻，就是先把数字拉出帽子。
您无需事先做到这一点。您还可以在给定当前板状态的情况下创建列表，然后从该列表中随机选择。
基本理念是......
提示：您可以使用单个值代表电路板单元......
Given a board[x][y] then...
cell = x * 3 + y
x = cell / 3
y = cell - x * 3