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']]

對于當(dāng)前需要點擊的點感凤，我們先判斷是不是雷悯周，是的話直接標(biāo)記X返回即可。如果不是的話陪竿，我們就數(shù)該點周圍的雷個數(shù)禽翼，如果周圍有雷，則當(dāng)前點變?yōu)槔椎膫€數(shù)并返回族跛。如果沒有的話闰挡，我們再對周圍所有的點調(diào)用遞歸函數(shù)再點擊即可。

This is a typical Search problem, either by using DFS or BFS. Search rules:
If click on a mine (’M’), mark it as ‘X’, stop further search.
If click on an empty cell (’E’), depends on how many surrounding mine:
2.1 Has surrounding mine(s), mark it with number of surrounding mine(s), stop further search.
2.2 No surrounding mine, mark it as ‘B’, continue search its 8 neighbors.

Solution1：DFS

思路：
Time Complexity: O(N) Space Complexity: O(N)

Solution2：BFS

思路：
Time Complexity: O(N) Space Complexity: O(N)

Solution1 Code：

public class Solution {
    public char[][] updateBoard(char[][] board, int[] click) {
        int m = board.length, n = board[0].length;
        int row = click[0], col = click[1];
        
        if (board[row][col] == 'M') { // Mine
            board[row][col] = 'X';
        }
        else { // Empty
            // Get number of mines first.
            int count = 0;
            for (int i = -1; i < 2; i++) {
                for (int j = -1; j < 2; j++) {
                    if (i == 0 && j == 0) continue;
                    int r = row + i, c = col + j;
                    if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
                    if (board[r][c] == 'M' || board[r][c] == 'X') count++;
                }
            }
            
            if (count > 0) { // If it is not a 'B', stop further DFS.
                board[row][col] = (char)(count + '0');
            }
            else { // Continue DFS to adjacent cells.
                board[row][col] = 'B';
                for (int i = -1; i < 2; i++) {
                    for (int j = -1; j < 2; j++) {
                        if (i == 0 && j == 0) continue;
                        int r = row + i, c = col + j;
                        if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
                        if (board[r][c] == 'E') updateBoard(board, new int[] {r, c});
                    }
                }
            }
        }
        
        return board;
    }
}

Solution2 Code：

public class Solution {
    public char[][] updateBoard(char[][] board, int[] click) {
        int m = board.length, n = board[0].length;
        Queue<int[]> queue = new LinkedList<>();
        queue.add(click);
        
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            int row = cell[0], col = cell[1];
            
            if (board[row][col] == 'M') { // Mine
                board[row][col] = 'X';
            }
            else { // Empty
                // Get number of mines first.
                int count = 0;
                for (int i = -1; i < 2; i++) {
                    for (int j = -1; j < 2; j++) {
                        if (i == 0 && j == 0) continue;
                        int r = row + i, c = col + j;
                        if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
                        if (board[r][c] == 'M' || board[r][c] == 'X') count++;
                    }
                }
                
                if (count > 0) { // If it is not a 'B', stop further BFS.
                    board[row][col] = (char)(count + '0');
                }
                else { // Continue BFS to adjacent cells.
                    board[row][col] = 'B';
                    for (int i = -1; i < 2; i++) {
                        for (int j = -1; j < 2; j++) {
                            if (i == 0 && j == 0) continue;
                            int r = row + i, c = col + j;
                            if (r < 0 || r >= m || c < 0 || c < 0 || c >= n) continue;
                            if (board[r][c] == 'E') {
                                queue.add(new int[] {r, c});
                                board[r][c] = 'B'; // Avoid to be added again.
                            }
                        }
                    }
                }
            }
        }
        
        return board;
    }
}