我写了一个程序的任务，我们必须写一个简单的五子棋程序。我以为我已经拥有了一切，但当我编译并运行时，它会引发胜利的场景，即使我只有4个同类，即使它们并不相邻。（只有在一种情况下，如果有五个连续的X或O，它才应该引发胜利）。我觉得我应该在每回合之前把计数器重置为0，但我不确定我应该在哪里做。任何提示将不胜感激！
import java.util.Scanner;
public class Gomoku1
{
public static void main (String[] args)
{
Scanner input = new Scanner(System.in);
char[][] map = new char [19][19];
int row = 0;
int column = 0;
//fill game with dots
for (int i = 0; i < map.length; i++)
{
for (int j = 0; j < map[i].length; j++)
{
map[i][j] = '.';
}
}
printMap(map);
char player1Choice = 'X';
char player2Choice = 'O';
int [] place;
while (true)
{
System.out.println("Player 1's turn!");
place = userTurn(map, player1Choice);
if (isValidMove(map, place[0], place[1]) == false)
{
System.out.println("Invalid move! Try again!");
place = userTurn(map, player1Choice);
}
if (isValidMove(map, place[0], place[1])) {
map[place[0]][place[1]] = player1Choice;
printMap(map);
}
if (isBoardFull(map) == true)
{
System.out.println("Board is full. Tied game.");
break;
}
if (hasPlayerWon(map, player1Choice) == true)
{
System.out.println("Player 1 Wins!");
break;
}
else
{
System.out.println("Player 2's turn!: ");
place = userTurn(map, player2Choice);
//System.out.println(isValidMove(map, row, column));
if (isValidMove(map, place[0], place[1]) == false)
{
System.out.println("Invalid move! Try again!");
place = userTurn(map, player2Choice);
}
if (isValidMove(map, place[0], place[1])) {
map[place[0]][place[1]] = player2Choice;
printMap(map);
}
if (isBoardFull(map) == true)
{
System.out.println("Board is full. Tied game.");
break;
}
if (hasPlayerWon(map, player2Choice) == true)
{
System.out.println("Player 2 Wins!");
break;
}
}
}
}
public static void printMap (char[][] map)
{
for (int i = 0; i < map.length; i++)
{
for (int j = 0; j < map[i].length; j++)
{
System.out.printf("%2c", map[i][j]);
}
System.out.println();
}
}
public static int [] userTurn (char[][] map, char playerChoice)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter row: ");
int row = input.nextInt();
System.out.print("Enter column: ");
int column = input.nextInt();
int place [] = {row, column};
return place;
}
public static boolean isValidMove (char[][] map, int row, int column)
{
//System.out.println ("n is valid move");
if (row < 0 || row > 18 || column < 0 || column > 18 || map[row][column]=='O' || map[row][column]=='X')
{
return false;
}
else
{
return true;
}
}
public static boolean isBoardFull (char[][] map)
{
int openSpots = 0;
for (int i = 0; i < map.length; i++)
{
for (int j = 0; j < map.length; j++)
{
if (!(map[i][j]=='.'))
openSpots++;
}
}
if (openSpots == 361)
{
return true;
}
return false;
}
public static boolean hasPlayerWon(char[][] map, int player)
{
if (isHorizontalWin(map, player) == true || isVerticalWin(map, player) == true || isDiagonalWin(map, player) == true)
{
return true;
}
return false;
}
public static boolean isHorizontalWin(char[][] map, int player)
{
int count = 0;
int r;
int c;
for (int i = 0; i < map.length; i++)
{
for (int j = 0; j < map.length; j++)
{
if (map[i][j]==(player))
{
r = i;
c = j;
while (r >= 0 && r <= 18 && c >= 0 && c <= 18 && map[r][c] == player)
{
count ++;
r += 0;
c += 1;
}
}
}
}
if (count == 5)
{
return true;
}
return false;
}
public static boolean isVerticalWin(char[][] map, int player)
{
int count = 0;
int r;
int c;
for (int i = 0; i < map.length; i++)
{
for (int j = 0; j < map.length; j++)
{
if (map[i][j]==(player))
{
r = i;
c = j;
while (r >= 0 && r <= 18 && c >= 0 && c <= 18 && map[r][c] == player)
{
count ++;
r += 1;
c += 0;
}
}
}
}
if (count == 5)
{
return true;
}
return false;
}
public static boolean isDiagonalWin(char[][] map, int player)
{
int count = 0;
int r;
int c;
for (int i = 0; i < map.length; i++)
{
for (int j = 0; j < map.length; j++)
{
if (map[i][j]==(player))
{
r = i;
c = j;
while (r >= 0 && r <= 18 && c >= 0 && c <= 18 && map[r][c] == player)
{
count++;
r += 1;
c += 1;
}
}
}
}
if (count == 5)
{
return true;
}
return false;
}
}
2条答案
按热度按时间rryofs0p1#
您在检查获胜条件的所有三个函数中都有问题：
isHorizontalWin、isVerticalWin和isDiagonalWin。这三个变量都递增变量count，但这个变量永远不会被设置回零。此外，检查count == 5是否应该在循环内进行。下面是如何修复isHorizontalWin的示例：ztyzrc3y2#
我知道已经有一段时间了...下面是我在任何方向上连续检测五个的尝试，但不是六个或更多的一部分。它是JavaScript，但你可以很容易地转换成Java。它更紧凑一些，在一个函数中完成所有工作，并带有两个辅助函数。它同时计算从x，y方向离开的两个方向上的连接石头。board[][]是一个简单的2D数组，Y值作为第一个索引。