Java 测试井字游戏获胜条件
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原文地址: http://stackoverflow.com/questions/18548265/
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testing tic tac toe win condition
提问by wjhplano
I'm looking for the most efficient javaway to test if somebody has won at tic tac toe. The data is in a 2d array like so...
我正在寻找最有效的Java方式来测试是否有人在井字游戏中获胜。数据在一个二维数组中，就像这样......
char[][] ticTacToe =
{{'X',' ','O'},
{'O','X','O'},
{'X',' ','X'},};
I know this isn't the professional way to initialize an array but I'm just testing here.
我知道这不是初始化数组的专业方法，但我只是在这里测试。
The best I can do for right now is an exhaustive if/else tree. Here's one of those trees...
我现在能做的最好的是一个详尽的 if/else 树。这是其中一棵树...
if (ticTacToe[1][1] == 'X'){
if (ticTacToe[0][0] == 'X'){
if (ticTacToe[2][2] == 'X'){
System.out.println("X wins");
}
}
else if (ticTacToe[0][1] == 'X'){
if (ticTacToe[2][1] == 'X'){
System.out.println("X wins");
}
}
else if (ticTacToe[1][0] == 'X'){
if (ticTacToe[1][2] == 'X'){
System.out.println("X wins");
}
}
else if (ticTacToe[2][0] == 'X'){
if (ticTacToe[0][2] == 'X'){
System.out.println("X wins");
}
}
}
This one only cares about what's in the middle
这个只关心中间的东西
This is very basic and I want to improve it as far as minimizing lines of code goes.
这是非常基本的，我想改进它，尽量减少代码行。
采纳答案by MrLore
It's a bit verbose, but I think this is probably the most efficient way to do it (unless someone can come up with a clever way to check both diagonals at once).
这有点冗长，但我认为这可能是最有效的方法（除非有人能想出一种聪明的方法同时检查两条对角线）。
public class TicTacToe
{
char[][] ticTacToe =
{{'X',' ','O'},
{'O','X','O'},
{'X',' ','X'},};
private Character winner = null;
public Character getWinner()
{
return this.winner;
}
public boolean isSolved()
{
this.checkSolved();
return this.winner != null;
}
private void checkSolved()
{
for(int i = 0; i < ticTacToe.length; i++)
{
Character win = checkRow(i);
if(win != null || (win = checkColumn(i)) != null)
{
this.winner = win;
return;
}
}
//Check diagonal top left to bottom right
if(this.ticTacToe[0][0] != ' ')
{
if(this.ticTacToe[0][0] == this.ticTacToe[1][1] &&
this.ticTacToe[1][1] == this.ticTacToe[2][2])
{
this.winner = this.ticTacToe[0][0];
}
}
//Check diagonal top right to bottom left
else if(this.ticTacToe[0][2] != ' ')
{
if(this.ticTacToe[0][2] == this.ticTacToe[1][1] &&
this.ticTacToe[1][1] == this.ticTacToe[2][0])
{
this.winner = this.ticTacToe[0][2];
}
}
}
private Character checkRow(int row)
{
if(this.ticTacToe[row][0] == ' ')
{
return null;
}
if(this.ticTacToe[row][0] == this.ticTacToe[row][1] &&
this.ticTacToe[row][1] == this.ticTacToe[row][2])
{
return this.ticTacToe[row][0];
}
return null;
}
private Character checkColumn(int column)
{
if(this.ticTacToe[0][column] == ' ')
{
return null;
}
if(this.ticTacToe[0][column] == this.ticTacToe[1][column] &&
this.ticTacToe[1][column] == this.ticTacToe[2][column])
{
return this.ticTacToe[column][0];
}
return null;
}
public static void main(String[] args)
{
TicTacToe ttt = new TicTacToe();
if(ttt.isSolved())
{
System.out.println(ttt.getWinner()); // X
}
}
}
回答by ???? ????
Just for fun, keep two numbers, starting as zeros, one for X, one for O. Update them by oring with the moves. To check for a winner, first and, then xorwith the mask.
只是为了好玩，保留两个数字，从零开始，一个是X，一个是O。通过or移动更新它们。要检查获胜者，首先and，然后xor使用面具。
277 & 273 ^ 273
0 ==> we have a winner.
276 & 273 ^ 273
1 ==> not.
277 == parseInt("100010101",2)
273 == parseInt("100010001",2)
276 == parseInt("100010100",2)
277 == parseInt("100010101",2)
273 == parseInt("100010001",2)
276 == parseInt("100010100",2)
For more fun, here's an example that plays Oin your favorite JavaScript console:
为了更有趣，这里有一个O在您最喜欢的 JavaScript 控制台中播放的示例：
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-type" content="text/html;charset=UTF-8">
</head>
<body>
<script>
var x = 0, o = 0, count = 0, w = 0
ws = [0007,0070,0700,0111,0222,0444,0124,0421]
function t1(v){
var w1 = 0
for (var i in ws)
w1 |= !(v & ws[i] ^ ws[i])
return w1
}
function t(i){
var ot = count % 2, m = 1 << (9 - i), bd = x | o
if (!ot && (i > 9 || i < 1 || i != Math.floor(i)))
return "Out of bounds."
else if (m & bd)
return "Position taken."
if (ot){
var n1 = 0, a1 = -2
while (bd & (1 << n1))
n1++
var n = n1
while (n1 < 9){
var m1 = 1 << n1
if (!(bd & m1)){
var bt = -mx(x,o | m1,count + 1)
if (bt > a1){
a1 = bt
n = n1
}
}
n1++
}
w = t1(o |= 1 << n)
}
else
w = t1(x |= m)
var b = "\n", p = 0400
while (p > 0){
if (p & x)
b += "X"
else if (p & o)
b += "O"
else b += "."
if (p & 0110)
b += "\n"
p >>= 1
}
if (w)
b += "\n\n" + (ot ? "O" : "X") + " wins!"
else if (!(bd ^ 0777))
b += "\n\nDraw."
if (!ot){
console.log(b + '\n\n"""')
count++
console.log(t(-1))
count++
}
else
return b + "\n"
return '"'
}
function mx(x1,o1,c1){
var ot1 = c1 % 2, w1 = ot1 ? t1(x1) : t1 (o1),
b1 = x1 | o1, p = 0400
if (w1)
return -1
if (!(b1 ^ 0777))
return 0
var a = -2
while (p > 0){
if (!(b1 & p))
a = Math.max(a,-mx(ot1 ? x1 : x1 | p,ot1 ? o1 | p : o1,c1 + 1))
p >>= 1
}
return a
}
console.log(' Plays O!'
+ '\nTo play, type t(MOVE); MOVE is from 1-9')
</script>
</body>
</html>
回答by Philipp
There are four different ways to win at tick-tack-toe:
有四种不同的方法可以在滴答声中获胜：
- form a horizontal line
- form a vertical line
- form a diagonal line from the upper-left to the lower-right corner
- form a diagonal line from the lower-left to the upper-right corner
- 形成一条水平线
- 形成一条垂直线
- 形成一条从左上角到右下角的对角线
- 形成一条从左下角到右上角的对角线
All of these four win-conditions can be solved with a for-loop. The advantage of this solution is that it can be applied to any matrix-size.
所有这四个获胜条件都可以通过 for 循环来解决。此解决方案的优点是它可以应用于任何矩阵大小。
回答by Yu-Han Lyu
For a player, say 'x', there are 8 ways to win and each corresponds to 3 'x' in a row/column/diagonal. Hence, you can create an array of length 8 and each item corresponds to the number of 'x' in that row/column/diagonal. When the player chooses a move, then you update the array and check whether there exists 3 in the array. Although it needs more space, it is easier to generalize to a large board.
对于玩家来说，比如说“x”，有 8 种获胜方式，每种方式对应于一行/列/对角线上的 3 个“x”。因此，您可以创建一个长度为 8 的数组，每个项目对应于该行/列/对角线中“x”的数量。当玩家选择移动时，然后更新数组并检查数组中是否存在 3。虽然它需要更多的空间，但更容易推广到大板。
回答by Marcin Szymczak
Mark board as 3x3 magicSquareand you have win when sum in line is 15.
将棋盘标记为 3x3 magicSquare，当在线总和为 15 时，您就赢了。
