LeetCode 529 扫雷游戏

扫雷游戏

class Solution {
public:
   int cal(int x, int y, vector>& board)
{
	int col = board.size();
	int row = board[0].size();
	int dx[8] = { 0, 1, 1, 1, 0, -1, -1, -1 };
	int dy[8] = { 1, 1, 0, -1, -1, -1, 0, 1 };
	int count = 0;
	for (int i = 0; i != 8; ++i)
	{
		int nx = x + dx[i], ny = y + dy[i];
		if (nx >= 0 && nx < col && ny >= 0 && ny < row)
		{
			if (board[nx][ny] == 'M')
				++count;
		}
	}
	return count;
}
void BFSa(vector>& board, int x, int y)
{
	int dx[8] = { 0, 1, 1, 1, 0, -1, -1, -1 };
	int dy[8] = { 1, 1, 0, -1, -1, -1, 0, 1 };
	int col = board.size();
	int row = board[0].size();
	int count_num = 0;
	count_num = cal(x, y, board);
	if (count_num == 0)//如果满足条件的话进入
	{
		board[x][y] = 'B';//相当于visit的条件改变
		for (int i = 0; i < 8; ++i)
		{
			int nx = x + dx[i];
			int ny = y + dy[i];
			if (nx >= 0 && nx < col&&ny >= 0 && ny < row)
			{
				if(board[nx][ny]=='E')//查看visit条件是什么
					BFSa(board, nx, ny);
			}
		}
	}
	else//不满足条件的话就退出
	{
		board[x][y] = count_num + '0';
		return;
	}
}
vector> updateBoard(vector>& board, vector& click) {
	//扫雷游戏和数独游戏差不多的,我们先选择深度优先遍历的算法
	int col = board.size();
	int row = board[0].size();
	int x = click[0];
	int y = click[1];
	if (board[x][y] == 'M')
	{
		board[x][y] = 'X';
		return board;
	}
	//如果一开始不是地雷需要宽度优先遍历进行相应的计算
	BFSa(board, x, y);
	return board;
}
};