Let’s play the minesweeper game (Wikipedia, online game)!

You are given a 2D char matrix representing the game board. ‘M’ represents an unrevealed mine, ‘E’ represents an unrevealed empty square, ‘B’ represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, digit (‘1’ to ‘8’) represents how many mines are adjacent to this revealed square, and finally ‘X’ represents a revealed mine.

Now given the next click position (row and column indices) among all the unrevealed squares (‘M’ or ‘E’), return the board after revealing this position according to the following rules:

If a mine (‘M’) is revealed, then the game is over - change it to ‘X’.
If an empty square (‘E’) with no adjacent mines is revealed, then change it to revealed blank (‘B’) and all of its adjacent unrevealed squares should be revealed recursively.
If an empty square (‘E’) with at least one adjacent mine is revealed, then change it to a digit (‘1’ to ‘8’) representing the number of adjacent mines.
Return the board when no more squares will be revealed.

Example 1:

Input:

[[‘E’, ‘E’, ‘E’, ‘E’, ‘E’],
[‘E’, ‘E’, ‘M’, ‘E’, ‘E’],
[‘E’, ‘E’, ‘E’, ‘E’, ‘E’],
[‘E’, ‘E’, ‘E’, ‘E’, ‘E’]]

Click : [3,0]

Output:

[[‘B’, ‘1’, ‘E’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘M’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘1’, ‘1’, ‘B’],
[‘B’, ‘B’, ‘B’, ‘B’, ‘B’]]

Explanation:

Example 2:

Input:

[[‘B’, ‘1’, ‘E’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘M’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘1’, ‘1’, ‘B’],
[‘B’, ‘B’, ‘B’, ‘B’, ‘B’]]

Click : [1,2]

Output:

[[‘B’, ‘1’, ‘E’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘X’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘1’, ‘1’, ‘B’],
[‘B’, ‘B’, ‘B’, ‘B’, ‘B’]]

Explanation:

Note:

1. The range of the input matrix’s height and width is [1,50].
2. The click position will only be an unrevealed square (‘M’ or ‘E’), which also means the input board contains at least one clickable square.
3. The input board won’t be a stage when game is over (some mines have been revealed).
4. For simplicity, not mentioned rules should be ignored in this problem. For example, you don’t need to reveal all the unrevealed mines when the game is over, consider any cases that you will win the game or flag any squares.

方法1: dfs

思路：

首先解决点到雷的可能，直接改成‘X’就可以返回了。否则要统计这一点周围有多少’M’，根据统计值，填入对应数字。如果count = 0，填入“B”，并且要向八个方向中**同样‘E’**的点发起dfs，这里E起到unvisited的标志作用，为了避免再向’B’循造。

class Solution {
public:
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        int m = board.size(), n = board[0].size();
        int row = click[0], col = click[1];
        if (board[row][col] == 'M') {
            board[row][col] = 'X';
        }
        else {
            int count = 0;
            for (int i = -1; i < 2; i++) {
                for (int j = -1; j < 2; j++) {
                    int x = row + i, y = col + j;
                    if (x < 0 || x >= m || y < 0 || y >= n) continue;
                    if (board[x][y] == 'M') count++;
                }
            }
            if (count) {
                board[row][col] = count + '0';
            }
            else {
                board[row][col] = 'B';
                for (int i = -1; i < 2; i++) {
                    for (int j = -1; j < 2; j++) {
                        int x = row + i, y = col + j;
                        if (x < 0 || x >= m || y < 0 || y >= n) continue;
                        if (board[x][y] == 'E') {
                            vector<int> next = {x, y};
                            updateBoard(board, next);
                        }
                    }
                }
            }
        }
        
        return board;
    }
};

方法2: bfs

思路：

class Solution {
public:
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        // BFS
        int m = board.size();
        if(0 == m) { return {}; }
        int n = board.front().size();
        if(0 == n) { return {}; }
        
        int cr = click[0];
        int cc = click[1];
        if(board[cr][cc] == 'M') { board[cr][cc] = 'X'; return board; }
        if(board[cr][cc] != 'E') { return {}; }
        
        vector<int> dx = {-1, -1, -1, 0,  0, 1, 1, 1};
        vector<int> dy = { 1,  0, -1, 1, -1, 1, 0, -1};
        auto get_mines = [&](int r, int c)->int{
            int cnt = 0;
            for(int i = 0; i < dx.size(); ++i)
            {
                int nx = dx[i] + r;
                int ny = dy[i] + c;
                if(0 <= nx && nx < m && 0 <= ny && ny < n && board[nx][ny] == 'M')
                { ++cnt; }
            }
            return cnt;
        };
        
        queue<pair<int,int>> q;
        q.push({cr, cc});
        board[cr][cc] = 'V';
        while(!q.empty())
        {
            auto cur = q.front(); q.pop();
            int r = cur.first, c = cur.second;
            int num_mines = get_mines(r, c);
            if(num_mines == 0)
            {
                board[r][c] = 'B';
                for(int i = 0; i < dx.size(); ++i)
                {
                    int nx = dx[i] + r;
                    int ny = dy[i] + c;
                    if(0 <= nx && nx < m && 0 <= ny && ny < n && board[nx][ny] == 'E')
                    {
                        q.push({nx, ny});
                        board[nx][ny] = 'V';
                    }
                }
            }
            else
            {
                board[r][c] = '0' + num_mines;
            }
        }
        return board;
    }
};