问题描述
首先我是初学Java编程。我想为学校项目的应用程序。我决定做一个棋盘游戏！ pretty的直线前进。！[董事会] [1] 这是董事会，并且有两个球员（人力）。第一个播放器（X）将他的举动在黑板上的任何地方。有提供给他9的地方。
first of all i'm beginner to Java programming. I want to make an application for School project. I have decided to make a board game! pretty straight forward.![Board][1] This is the board and there are two player (Both Human). first one player(x) will make his move anywhere on the board. there are 9 places available to him.
本场比赛的动机是把所有三个标记在一行对角，水平或垂直。
motive of the game is to put all three tokens in one line diagonally, horizontally or vertically.
！[获奖情况] [2]我想算法检查赢家，如何挑选价值FRM用户时，他让一招。我浏览大量的井字TOH算法，但我不知道该怎么去与他们错了。！[更多钞票移动] [3]请帮忙！！
![Winning condition][2]i want algorithm to check the winner and how to pick value frm user when he makes a move.i browsed lots of tic tac toh algorithm but i dont know what goes wrong with them.![Posible moves][3]Please Help!!
推荐答案
由于它的功课，我不想只是给你一个答案。此外，你没有什么你都在努力是很清楚，所以我希望这是适用于您遇到的（S）的问题。编辑你的答案，使之更加清晰和/或发表评论，让所有谁回答可能会提高他们的答案更有效地满足您的需求。的
您可以使用一个二维数组的存储板。在Java中，语法用于创建一个3x3的二维数组（例如，使用字符类型，存储 X 或 0 ）如下：
You can make use of a two-dimensional array to store the board. In Java, the syntax for creating a 3x3 two-dimensional array (e.g. using the char type, to store X or O) is as follows:
char[][] board = new char[3][3];
在井字棋，有8种可能板上规定，表明玩家在游戏中获胜（标记 X 下面的动作）：
In Tic-Tac-Toe, there are 8 possible board states that indicate that a player has won the game (moves marked with X below):
(1) (2) (3) (4) (5) (6) (7) (8)
X X X # # # # # # X # # # X # # # X X # # # # X
# # # X X X # # # X # # # X # # # X # X # # X #
# # # # # # X X X X # # # X # # # X # # X X # #
记住，数组下标0开始（并上升到2，因为我们有一个3x3的二维数组），这相当于下面的一组条件：
Remembering that array indices start at 0 (and go up to 2, since we have a 3x3 2D array), this translates to the following set of conditions:
(1): board[0][0] == board[0][1] && board[0][1] == board[0][2]
(2): board[1][0] == board[1][1] && board[1][1] == board[1][2]
(3): board[2][0] == board[2][1] && board[2][1] == board[2][2]
(4): board[0][0] == board[1][0] && board[1][0] == board[2][0]
(5): board[0][1] == board[1][1] && board[1][1] == board[2][1]
(6): board[0][2] == board[1][2] && board[1][2] == board[2][2]
(7): board[0][0] == board[1][1] && board[1][1] == board[2][2]
(8): board[0][2] == board[1][1] && board[1][1] == board[2][0]
如果任何这些细胞是相等的（并且不为空），然后有人（无论是 X 或 0 ，这取决于谁占有这些细胞）已经赢得了比赛。你可能使用循环，使这个code更紧凑。
If any of these cells are equal (and are not blank) then someone (either X or O, depending on who occupies those cells) has won the game. You could potentially use loops to make this code more compact.
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