Question

我已经获得了一个tic tac toe游戏的代码。我制作了代码来检查垂直是否会赢，并试图检查对角线。我能够检查主对角线，但似乎无法确定如何检查辅助对角线。我认为我的代码可以工作，但事实并非如此。我遇到的问题从第172行开始

var array_nums = Array.apply(null, Array(50)).map(function() { 
    return Math.floor(Math.random() * 50);
}).reduce(function (acc, curr) {
    if (curr >= 10 && curr <= 19) {
        acc[curr] = (acc[curr] || 0) + 1;
        acc["all"]++;
    }
    return acc;
}, {all:0});

document.getElementById("results").innerHTML = JSON.stringify(array_nums);

Answer 1

您的代码逻辑错误。如果主对角线上的单元格不等于mark，则只检查辅助对角线。

您需要两个单独的变量来跟踪每条对角线是否有胜利。您的代码应如下所示：

int match_prime = 1, match_second = 1;
for(col = 0;col < BOARD_SIZE;++col){
    match_prime = board[col][col] == mark;
    match_second = board[BOARD_SIZE - col - 1][col] == mark;
}
won = match_prime || match_second;

Answer 2

方法1

对角元素索引的硬编码。

int hasWonDiagonal(char board[BOARD_SIZE][BOARD_SIZE], char mark) {
   int won = 0;
   int match = 0;

   if ( ( board[0][0] == mark &&
          board[1][1] == mark &&
          board[2][2] == mark ) &&
        ( board[0][2] == mark &&
          board[1][1] == mark &&
          board[2][0] == mark ) )

   {
      match = 1;
   }
   won = match;

   return won; // Stub -- make this return the correct value
}

方法2

使用for循环并迭代索引。

int hasWonDiagonal(char board[BOARD_SIZE][BOARD_SIZE], char mark) {
   int match = 1;
   int won = 0;
   int row = 0;

   // Check the first diagonal.
   for (row = 0; row < BOARD_SIZE && !won; row++) {
      if(board[row][row] != mark) {
         match=0;
      }
   } 

   // If the first diagonal check already produces a match,
   // there is no need to check the second diagonal.
   if ( match != 1 )
   {
      int col = BOARD_SIZE-1;
      for (row = 0; row < BOARD_SIZE && !won; row++, col--) {
         if(board[row][col] != mark){
            match=0;
         }  
      } 
   }
   won = match;

   return won; // Stub -- make this return the correct value
}

Answer 3

这些功能可以采用以下方式

//
// Determines if the 'mark' completely fills a row
// returns 1 if yes, 0 if no
//
int hasWonHorizontal( char board[BOARD_SIZE][BOARD_SIZE], char mark ) 
{
    int won = 0; // boolean (0/1).  

    for ( int row = 0; row < BOARD_SIZE && !won; row++ ) 
    {
        int col = 0;

        while ( col < BOARD_SIZE && board[row][col] == mark ) ++col 

        won = col == BOARD_SIZE;
    } 

    return won;
}


//
// Determines if the 'mark' completely fills a column
// returns 1 if yes, 0 if no
//
int hasWonVertical( char board[BOARD_SIZE][BOARD_SIZE], char mark ) 
{
    int won = 0;

    for ( int col = 0; col < BOARD_SIZE && !won; col++ ) 
    {
        int row = 0;

        while ( row < BOARD_SIZE && board[row][col] == mark ) ++row;

        won = row == BOARD_SIZE;     
    }

    return won; // Stub -- make this return the correct value
}


//
// Determines if the 'mark' completely fills a diagonal
// returns 1 if yes, 0 if no
//
int hasWonDiagonal(char board[BOARD_SIZE][BOARD_SIZE], char mark) 
{
    int won = 0;
    int i = 0;

    while ( i < BOARD_SIZE && board[i][i ] == mark ) ++i;

    won = i == BOARD_SIZE;

    if ( !won )
    {
        i = 0;
        while ( i < BOARD_SIZE && board[i][BOARD_SIZE - i - 1 ] == mark ) ++i;
        won = i == BOARD_SIZE;
    }

    return won; // Stub -- make this return the correct value
}

检查对角线是否在tic tac toe中获胜

3 个答案: