我正在为nxn板上的n个玩家制作一个tic tac toe游戏，但获胜条件是连续3个。我到目前为止解决这个问题的方法是：当一个移动时，程序将检查以下方块是否连续3个。
(x-1,y+1) (x,y+1) (x+1,y+1)
(x-1,y) (x,y) (x+1,y)
(x-1,y-1) (x,y-1) (x+1,y-1)
它将检查顶部（x-1，y + 1）（x，y + 1）（x + 1，y + 1）底部（x-1，y-1）（x，y-1） （X + 1，Y-1） 边（x + 1，y + 1）（x + 1，y）（x + 1，y-1），（x-1，y + 1）（x-1，y）（x-1，y- 1），对角线和穿过中间（x，y）的对角线。
到目前为止，我的代码是： public int checkWinning() {
for(int a = 1; a < size-1; a++){
for(int b = 1; b < size-1; b++){
if (board[a][b] == board[a+1][b] && board[a][b] == board[a-1][b]){
return board[a][b];
}else if(board[a][b] == board[a][b+1] && board[a][b] == board[a][b-1]){
return board[a][b];
}else if(board[a][b] == board[a+1][b-1] && board[a][b] == board[a-1][b+1]){
return board[a][b];
}else if(board[a][b] == board[a+1][b+1] && board[a][b] == board[a-1][b-1]){
return board[a][b];
}
}
}
for(int d = 1; d < size-1; d++){
if (board[0][d] == board[0][d-1] && board[0][d] == board[0][d+1]){
return board[0][d];
} else if (board[size-1][d] == board[size-1][d-1] && board[size-1][d] == board[size-1][d+1]){
return board[size-1][d];
}
}
for(int c = 1; c < size-1; c++){
if (board[c][0] == board[c-1][0] && board[c][0] == board[c+1][0]){
return board[c][0];
}else if(board[c][size-1] == board[c-1][size-1] && board[c][size-1] == board[c+1][size-1]){
return board[c][size-1];
}
}
return 0;
}
第一部分是我通过中间和对角线检查的部分。第二部分我检查顶部的底部和顶部和第三部分检查两侧。
当它返回0表示还没有赢家。
@override
public void checkResult() {
int winner = this.board.checkWinning();
if (winner > 0) {
this.ui.showResult("Player "+winner+" wins!");
}
if (this.board.checkFull()) {
this.ui.showResult("This is a DRAW!");
}
}
Board [x] [y] - ＆gt;表示板的二维数组，坐标从左上角（0,0）到右下角（大小-1，大小-1）计数，板[x] [y] == 0表示位置自由（ x，y），board [x] [y] == i for i＆gt; 0代表玩家我在（x，y）上移动，只是你知道它。
我的问题是，当我将电路板扩展到大于3x3的大小时，程序会以某种方式自行覆盖它，或者每次都不会检查每个方面的顶部和底部，而且我看起来似乎不太原因。
答案 0 :(得分：3)
编辑：
在应用程序中播放了几分钟...有趣的结果
java -jar tic-tac-toe.jar 5 20
It was a cats game!!
|1|1|5|5|1|3|5|3|1|5|2|5|1|1|2|
|2|3|2|3|1|5|3|5|3|2|3|1|5|2|2|
|5|4|5|4|1|5|5|4|2|1|4|5|4|2|2|
|3|2|1|5|5|5|2|4|5|3|4|1|2|4|2|
|3|4|1|2|5|4|1|1|4|5|1|3|3|4|1|
|1|5|4|4|3|2|5|1|3|5|1|3|5|3|4|
|2|5|1|4|3|3|3|5|3|1|1|4|3|4|4|
|1|4|5|1|1|5|4|5|2|4|1|1|5|4|3|
|1|3|2|1|4|2|4|3|3|4|5|2|4|3|3|
|5|1|1|3|3|4|4|4|2|2|1|4|3|2|5|
|2|2|3|1|5|5|4|1|3|5|3|2|3|3|2|
|2|4|2|4|4|1|3|1|1|3|1|2|1|2|2|
|2|5|5|1|4|3|4|5|5|4|5|3|3|5|2|
|4|5|2|1|5|3|2|1|3|2|2|2|2|4|4|
|4|1|1|4|5|4|5|4|2|2|3|3|2|2|3|
Played 100 games:
Number wins by Player1: 0
Number wins by Player2: 0
Number wins by Player3: 0
Number wins by Player4: 0
Number wins by Player5: 0
Number of ties: 100
没有滚动浏览所有100场比赛以找到获胜的棋盘，但我认为这很有趣：
java -jar tic-tac-toe.jar 2 10
Player2 won the game!
|1|1|2|1|2|2| |2|1|2|
|2|2|2|2|2|2|2|2|2|2|
|2|1|2|2|2|1|1|1|1|1|
|1|1|1|1|2|1|2|1|1|1|
|2|2| |1|2|1|1|1|1|2|
|2|2|2|1|1|1| |1|2|2|
|2|2|1|2|2|2|2|2|1|1|
| | |2|2|2|2| |1|1|1|
|1|1|2|2|2|1|1|1|1| |
| | |1|1|1|1|1|2|1| |
Played 100 games:
Number wins by Player1: 0
Number wins by Player2: 1
Number of ties: 99
这确实回答了你的问题...但我把它拿了一点......决定实施解决方案。 而不是计算匹配...我只是从最后一个玩家玩的点开始检查，如果行列中的所有标记和诊断与玩家匹配，他就赢了。
package com.clinkworks.example;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class TicTacToe {
private static final String TIE = "TIE";
private static final Map<String, Integer> gamesToWinsMap = new HashMap<String, Integer>();
/**
* accepts input in the following format:
*
* playerCount rowCount columnCount (sets the game with the n players, n columns, and n rows)
* - java -jar tic-tac-toe.jar 2 3 3
* PlayerCount squareSize (defaults to a game with rows and cols the same as squareSize and the player count given)
* - java -jar tic-tac-toe.jar 2 3
* PlayerCount (defaults to a 3 by 3 game)
* - java -jar tic-tac-toe.jar 2
* no input (defaults to a 3 by 3 game with 2 players)
* - java -jar tic-tac-toe.jar
* @param args
*/
public static void main(String[] args) {
int playerCount = 2;
int rows = 3;
int cols = 3;
if(args.length == 3){
playerCount = Integer.valueOf(args[0]);
rows = Integer.valueOf(args[1]);
cols = Integer.valueOf(args[2]);
}
if(args.length == 2){
playerCount = Integer.valueOf(args[0]);
rows = Integer.valueOf(args[1]);
cols = rows;
}
if(args.length == 1){
playerCount = Integer.valueOf(args[0]);
}
for(int i = 1; i <= playerCount; i++){
gamesToWinsMap.put("Player" + i, 0);
}
//lets play 100 games and see the wins and ties
playGames(100, playerCount, rows, cols);
for(int i = 1; i <= playerCount; i++){
System.out.println("Number wins by Player" + i + ": " + gamesToWinsMap.get("Player" + i));
}
System.out.println("Number of ties: " + gamesToWinsMap.get(TIE));
}
public static void playGames(int gamesToPlay, int playerCount, int rows, int cols) {
//play a new game each iteration, in our example, count = 100;
for (int i = 0; i < gamesToPlay; i++) {
playGame(playerCount, rows, cols);
}
}
public static void playGame(int playerCount, int rows, int cols) {
//create a new game board. this initalizes our 2d array and lets the complexity of handling that
// array be deligated to the board object.
Board board = new Board(playerCount, rows, cols);
//we are going to generate a random list of moves. Heres where we are goign to store it
List<Move> moves = new ArrayList<Move>();
//we are creating moves for each space on the board.
for (int row = 0; row < rows; row++) {
for (int col = 0; col < cols; col++) {
moves.add(new Move(row, col));
}
}
//randomize the move list
Collections.shuffle(moves);
//do each move
for (Move move : moves) {
board.play(move);
if(gameOver(board)){
break;
}
}
}
public static boolean gameOver(Board board){
if (board.whoWon() != null) {
System.out.println(board.whoWon() + " won the game!");
System.out.println(board);
Integer winCount = gamesToWinsMap.get(board.whoWon());
winCount = winCount == null ? 1 : winCount + 1;
gamesToWinsMap.put(board.whoWon(), winCount);
return true;
} else if (board.movesLeft() == 0) {
System.out.println("It was a cats game!!");
System.out.println(board);
Integer tieCount = gamesToWinsMap.get(TIE);
tieCount = tieCount == null ? 1 : tieCount + 1;
gamesToWinsMap.put(TIE, tieCount);
return true;
}
return false;
}
public static class Move {
private int row;
private int column;
public Move(int row, int column) {
this.row = row;
this.column = column;
}
public int getRow() {
return row;
}
public int getColumn() {
return column;
}
}
public static class Board {
private final int rowSize;
private final int columnSize;
private final Integer[][] gameBoard;
private int playerCount;
private int currentPlayer;
private String winningPlayer;
public Board() {
gameBoard = new Integer[3][3];
currentPlayer = 1;
winningPlayer = null;
this.rowSize = 3;
this.columnSize = 3;
playerCount = 2;
}
public Board(int players) {
gameBoard = new Integer[3][3];
currentPlayer = 1;
winningPlayer = null;
this.rowSize = 3;
this.columnSize = 3;
playerCount = players;
}
public Board(int rowSize, int columnSize) {
gameBoard = new Integer[rowSize][columnSize];
currentPlayer = 1;
winningPlayer = null;
playerCount = 2;
this.rowSize = rowSize;
this.columnSize = columnSize;
}
public Board(int players, int rowSize, int columnSize) {
gameBoard = new Integer[rowSize][columnSize];
currentPlayer = 1;
winningPlayer = null;
playerCount = players;
this.rowSize = rowSize;
this.columnSize = columnSize;
}
/**
*
* @return the amount of empty spaces remaining on the game board, or if theres a winning player, zero.
*/
public int movesLeft() {
if(whoWon() != null){
return 0;
}
int moveCount = 0;
for (int x = 0; x < getRowSize(); x++) {
for (int y = 0; y < getColumnSize(); y++) {
moveCount += getMoveAt(x, y) == null ? 1 : 0;
}
}
return moveCount;
}
/**
* If someone won, this will return the winning player.
*
* @return the winning player
*/
public String whoWon() {
return winningPlayer;
}
/**
* This move allows the next player to choose where to place their mark.
*
* @param row
* @param column
* @return if the game is over, play will return true, otherwise false.
*/
public boolean play(Move move) {
if (!validMove(move)) {
// always fail early
throw new IllegalStateException("Player " + getCurrentPlayer() + " cannot play at " + move.getRow() + ", " + move.getColumn() + "\n" + toString());
}
doMove(move);
boolean playerWon = isWinningMove(move);
if (playerWon) {
winningPlayer = "Player" + getCurrentPlayer();
return true;
}
shiftPlayer();
boolean outOfMoves = movesLeft() <= 0;
return outOfMoves;
}
public int getRowSize() {
return rowSize;
}
public int getColumnSize() {
return columnSize;
}
public int getCurrentPlayer() {
return currentPlayer;
}
public Integer getMoveAt(int row, int column) {
return gameBoard[row][column];
}
private void doMove(Move move) {
gameBoard[move.getRow()][move.getColumn()] = getCurrentPlayer();
}
private void shiftPlayer() {
if(getCurrentPlayer() == getPlayerCount()){
currentPlayer = 1;
}else{
currentPlayer++;
}
}
private int getPlayerCount() {
return playerCount;
}
private boolean validMove(Move move) {
boolean noMoveAtIndex = false;
boolean indexesAreOk = move.getRow() >= 0 || move.getRow() < getRowSize();
indexesAreOk = indexesAreOk && move.getColumn() >= 0 || move.getColumn() < getColumnSize();
if (indexesAreOk) {
noMoveAtIndex = getMoveAt(move.getRow(), move.getColumn()) == null;
}
return indexesAreOk && noMoveAtIndex;
}
private boolean isWinningMove(Move move) {
// since we check to see if the player won on each move
// we are safe to simply check the last move
return winsDown(move) || winsAcross(move) || winsDiagnally(move);
}
private boolean winsDown(Move move) {
boolean matchesColumn = true;
for (int i = 0; i < getColumnSize(); i++) {
Integer moveOnCol = getMoveAt(move.getRow(), i);
if (moveOnCol == null || getCurrentPlayer() != moveOnCol) {
matchesColumn = false;
break;
}
}
return matchesColumn;
}
private boolean winsAcross(Move move) {
boolean matchesRow = true;
for (int i = 0; i < getRowSize(); i++) {
Integer moveOnRow = getMoveAt(i, move.getColumn());
if (moveOnRow == null || getCurrentPlayer() != moveOnRow) {
matchesRow = false;
break;
}
}
return matchesRow;
}
private boolean winsDiagnally(Move move) {
// diagnals we only care about x and y being teh same...
// only perfect squares can have diagnals
// so we check (0,0)(1,1)(2,2) .. etc
boolean matchesDiagnal = false;
if (isOnDiagnal(move.getRow(), move.getColumn())) {
matchesDiagnal = true;
for (int i = 0; i < getRowSize(); i++) {
Integer moveOnDiagnal = getMoveAt(i, i);
if (moveOnDiagnal == null || moveOnDiagnal != getCurrentPlayer()) {
matchesDiagnal = false;
break;
}
}
}
return matchesDiagnal;
}
private boolean isOnDiagnal(int x, int y) {
if (boardIsAMagicSquare()) {
return x == y;
} else {
return false;
}
}
private boolean boardIsAMagicSquare() {
return getRowSize() == getColumnSize();
}
public String toString() {
StringBuffer stringBuffer = new StringBuffer();
for(int y = 0; y < getColumnSize(); y++) {
for(int x = 0; x < getRowSize(); x++) {
Integer move = getMoveAt(x, y);
String moveToPrint = "";
if (move == null) {
moveToPrint = " ";
} else {
moveToPrint = move.toString();
}
stringBuffer.append("|").append(moveToPrint);
}
stringBuffer.append("|\n");
}
return stringBuffer.toString();
}
}
}
答案 1 :(得分：0)
我必须修改我的答案。如果你想要连续三个，无论你的电路板大小如何，你的循环代码可能就足够了，但是你总是在检查字段的值是否相同但是从来没有区别和非空字段。
所以“空”也可以获胜，这将有效地隐藏玩家的可能胜利。换句话说，即使字段大小为3，您的代码也无法正常工作。你没有对它进行足够的测试。
如果我将电路板初始化为
int[][] board={
{ 1, 1, 1 },
{ 0, 0, 0 },
{ 0, 0, 0 },
};
您的代码返回0，因为第二行包含三个零。我假设0表示空字段，但“空”的实际值无关紧要。您必须从三行检查中排除空字段。
答案 2 :(得分：0)
你可以通过稍微打破逻辑来简化这个数量。
首先要意识到你只需要检查你刚放置的那个部分的胜利。
现在我们需要一种方法来检查这一举动是否是赢家。
首先，我们需要一个简单的函数来检查一个单元格是否与给定值匹配，如果它在边界内并匹配则返回true。
private boolean cellMatches(int x, int y, int val) {
if (x<0||x>boardWidth)
return false;
if (y<0||y>boardHeight)
return false;
return board[x][y]==val;
}
现在是一个函数，你给出一个起始位置（x和y）和一个delta（dx，dy），它会在该方向上检查最多两个单元格，返回一行中匹配值的数量。对于两次检查，for循环可能有点过分，但它可以很容易地扩展到更长的行使用。
private int countMatches(int x, int y, int dx, int dy, int val) {
int count = 0;
for (int step=1;step<=2;step++) {
if (cellMatches(x+dx*step, y+dy*step, val) {
count++;
} else {
return count;
}
}
return count;
}
现在我们可以使用前面的方法了。当我们放置一个新作品时，我们可以在每个匹配的方向上计算出来。合计计数是连续的总数。 （即连续两个顶部+ 1个机器人=总运行长度为4）。如果这些运行长度中的任何一个是3，那么这是一个成功的举动。
private boolean makeMove(int x, int y, int val) {
board[x][y] = val;
int runlength=countMatches(x,y,0,1,val) + countMatches(x,y,0,-1,val);
if (runLength >= 2)
return true;
int runlength=countMatches(x,y,1,0,val) + countMatches(x,y,-1,0,val);
if (runLength >= 2)
return true;
int runlength=countMatches(x,y,1,1,val) + countMatches(x,y,-1,-1,val);
if (runLength >= 2)
return true;
int runlength=countMatches(x,y,1,-1,val) + countMatches(x,y,-1,1,val);
if (runLength >= 2)
return true;
return false;
}
请注意，因为我们需要计算我们放置的中心部分，所以我们只需要两个或更多的运行长度。