题目思路：

代码设计：

代码呈现：

class Solution {int m,n;int[] dx = {0,0,-1,1,-1,-1,1,1};int[] dy = {-1,1,0,0,-1,1,-1,1};public char[][] updateBoard(char[][] board, int[] click) {m = board.length;n = board[0].length;int x = click[0], y = click[1];//直接点到地雷if(board[x][y] == 'M'){board[x][y] = 'X';return board;}dfs(board,x,y);return board;    }private void dfs(char[][] board, int i, int j){//统计周围地雷的个数int count = 0;for(int k = 0; k < 8; k++){int x = i + dx[k];int y = j + dy[k];if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M'){count++;}}//去遍历未挖出的空方块//周围没有地雷if(count == 0){board[i][j] = 'B';for(int k = 0; k < 8; k++){int x = i + dx[k];int y = j + dy[k];if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E'){dfs(board,x,y);}}}else {//有地雷：修改它周围地雷个数，并返回board[i][j] = (char)(count + '0');return;}}
}