这是一个DFS，虽然跟常规的有点差别，理解不是问题。B站的小姐姐好厉害！
class Solution {
private int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
public char[][] updateBoard(char[][] board, int[] click) {
/* ternimator */
if (board.length == 0 || click.length != 2)
return board;
int m = board.length, n = board[0].length;
int row = click[0], col = click[1];
/* process current logic */
// 1-如果是M或者X，变成X直接退出
if (board[row][col] == 'M' || board[row][col] == 'X') {
board[row][col] = 'X';
return board;
}
// 2-计算周边M的数量，如果有雷，更新数字并退出
int num = 0;
for (int[] dir : dirs) {
int newRow = dir[0] + row;
int newCol = dir[1] + col;
if (newRow >= 0 && newCol >= 0 && newRow < m && newCol < n && board[newRow][newCol] == 'M') {
num += 1;
}
}
if (num > 0) {
board[row][col] = (char)(num + '0');
return board;
} else {
board[row][col] = 'B';
}
/* drill down */
// 3-揭开周边的E,dfs
for (int[] dir : dirs) {
int newRow = dir[0] + row;
int newCol = dir[1] + col;
if (newRow >= 0 && newCol >= 0 && newRow < m && newCol < n && board[newRow][newCol] == 'E') {
updateBoard(board, new int[]{newRow, newCol});
}
}
return board;
}
}