我正在尝试在C中重新创建游戏2048，但我无法将算法移动或合并到一起以正常运行。 在最初的2048游戏中，您可以像这样将瓷砖移动到一起：
2 | 2 | 4 | 4 4 | 8 | |
---+---+---+--- *swipes to the left* -> ---+---+---+---
8 | | 8 | 16| | |
因此，两个相同的图块可以合并为一个尺寸为两倍的图块。我的版本几乎相同，但不是使用数字，而是使用在合并时递增1的字符，因此[A|A]将合并到[B]等等。我这样做只是为了不必处理不同大小的瓷砖。
所以我的电路板存储为一个4 * 4字符数组存在于一个名为grid的结构中（我知道可能有点冗余）
typedef struct grid {
char tiles[4][4];
} Grid;
我曾试图让算法向上，向下，向左和向右移动和合并，但它们无法正常工作。
void pushLeft(Grid * grid)
{
int i, j, k;
for(i = 0; i < 4; i++) //Row number i
{
for(j = 1; j < 4; j++) //Column number j
{
if(grid->tiles[i][j] != ' ') //tile is not empty
{
int flag = 1; //flag to prevent merging more than one level at a time
//Starting on column k, push tile as far to the left as possible
for(k = j; k > 0; k--)
{
if(grid->tiles[i][k-1] == ' ') //neighbor tile is empty
{
grid->tiles[i][k-1] = grid->tiles[i][k];
grid->tiles[i][k] = ' ';
}
else if(grid->tiles[i][k-1] == grid->tiles[i][k] && flag) //neighbor equals
{
grid->tiles[i][k-1]++;
grid->tiles[i][k] = ' ';
flag = 0;
}
else //Can't push or merge
{
flag = 1;
break;
}
}
}
} // Done with row
}
}
void pushRight(Grid * grid)
{
int i, j, k;
for(i = 0; i < 4; i++) //Row number i
{
for(j = 2; j >= 0; j--) //Column number j
{
if(grid->tiles[i][j] != ' ') //tile is not empty
{
int flag = 1; //flag to prevent merging more than one level at a time
//Starting on column k, push tile as far to the right as possible
for(k = j; k < 3; k++)
{
if(grid->tiles[i][k+1] == ' ') //neighbor tile is empty
{
grid->tiles[i][k+1] = grid->tiles[i][k];
grid->tiles[i][k] = ' ';
}
else if(grid->tiles[i][k+1] == grid->tiles[i][k] && flag) //neighbor equals
{
grid->tiles[i][k+1]++;
grid->tiles[i][k] = ' ';
flag = 0;
}
else //Can't push or merge
{
flag = 1;
break;
}
}
}
} // Done with row
}
}
void pushUp(Grid * grid)
{
int i, j, k;
for(i = 0; i < 4; i++) //Column number i
{
for(j = 1; j < 4; j++) //Row number j
{
if(grid->tiles[j][i] != ' ') //tile is not empty
{
int flag = 1; //flag to prevent merging more than one level at a time
//Starting on row k, push tile as far upwards as possible
for(k = j; k > 0; k--)
{
if(grid->tiles[k-1][i] == ' ') //neighbor tile is empty
{
grid->tiles[k-1][i] = grid->tiles[i][k];
grid->tiles[k][i] = ' ';
}
else if(grid->tiles[k-1][i] == grid->tiles[i][k] && flag) //neighbor equals
{
grid->tiles[k-1][i]++;
grid->tiles[k][i] = ' ';
flag = 0;
}
else //Can't push or merge
{
flag = 1;
break;
}
}
}
} // Done with column
}
}
void pushDown(Grid * grid)
{
int i, j, k;
for(i = 0; i < 4; i++) //Column number i
{
for(j = 2; j >= 0; j--) //Row number j
{
if(grid->tiles[j][i] != ' ') //tile is not empty
{
int flag = 1; //flag to prevent merging more than one level at a time
//Starting on row k, push tile as far down as possible
for(k = j; k < 3; k++)
{
if(grid->tiles[k+1][i] == ' ') //neighbor tile is empty
{
grid->tiles[k+1][i] = grid->tiles[i][k];
grid->tiles[k][i] = ' ';
}
else if(grid->tiles[k+1][i] == grid->tiles[i][k] && flag) //neighbor equals
{
grid->tiles[k+1][i]++;
grid->tiles[k][i] = ' ';
flag = 0;
}
else //Can't push or merge
{
flag = 1;
break;
}
}
}
} // Done with column
}
}
我用一些硬编码的测试数据测试了这些算法。将磁贴推向左侧的算法似乎正常工作。 pushRight几乎可以工作，但它同时合并两个级别，因此[B|A|A]合并到[C]但应合并到[B|B]。
pushUp似乎几乎总是用空的瓷砖（空格）擦拭整个电路板。 pushDows似乎正在移除一些瓷砖。
有没有人看到问题或知道这样做的方法？我曾考虑使用递归算法，但我无法绕过它。
答案 0 :(得分：0)
我会亲自将滑动拆分为两个步骤，因为向左滑动并向右滑动实际上在功能上与磁贴组合相同。唯一的区别是剩余的瓷砖根据方向聚集在左侧或右侧。
下面是一个快速算法，用一个新的瓷砖替换两个瓷砖。我向左扫描＆gt;向右扫描并用新的图块替换左图块，将右图块置零，然后确保从比较中排除这个新图块：
typedef struct grid {
char tiles[4][4];
} Grid;
void eliminateHoriz (Grid* g)
{
int row, col, col2;
for (row=0; row<4; row++)
{
for (col=0; col<4; col++)
{
if (g->tiles[row][col])
{
for (col2=col+1; col2<4; col2++)
{
if (g->tiles[row][col2])
{
if (g->tiles[row][col] == g->tiles[row][col2])
{
g->tiles[row][col++] *= 2;
g->tiles[row][col2] = 0;
}
break;
}
}
}
}
}
}
void showGrid (Grid* g)
{
int row, col;
for (row=0; row<4; row++)
for (col=0; col<4; col++)
printf ("%4d%c",
g->tiles[row][col],
col == 3 ? '\n' : ' ');
printf ("\n");
}
int main()
{
Grid g = {{2,2,4,4,
8,0,8,0,
8,8,8,4,
2,2,2,2}};
showGrid (&g);
eliminateHoriz (&g);
showGrid (&g);
system ("pause");
return 0;
}
输出：
2 2 4 4
8 0 8 0
8 8 8 4
2 2 2 2
4 0 8 0
16 0 0 0
16 0 8 4
4 0 4 0
在此之后，可以进行简单的压缩步骤，或者实时输出到第二个缓冲区，或者哪个。减少重复。