让我们一起来玩扫雷游戏！
给你一个大小为m x n二维字符矩阵board，表示扫雷游戏的盘面，其中：
'M'代表一个 未挖出的 地雷， 'E'代表一个 未挖出的 空方块， 'B'代表没有相邻（上，下，左，右，和所有4个对角线）地雷的 已挖出的 空白方块， 数字（'1' 到'8'）表示有多少地雷与这块 已挖出的 方块相邻， 'X'则表示一个 已挖出的 地雷。 给你一个整数数组click，其中click = [clickr, clickc]表示在所有未挖出的方块（'M'或者'E'）中的下一个点击位置（clickr是行下标，clickc是列下标）。
根据以下规则，返回相应位置被点击后对应的盘面：
如果一个地雷（'M'）被挖出，游戏就结束了- 把它改为'X'。 如果一个 没有相邻地雷 的空方块（'E'）被挖出，修改它为（'B'），并且所有和其相邻的 未挖出 方块都应该被递归地揭露。 如果一个 至少与一个地雷相邻 的空方块（'E'）被挖出，修改它为数字（'1'到'8'），表示相邻地雷的数量。 如果在此次点击中，若无更多方块可被揭露，则返回盘面。
示例 1：
输入：board = [["E","E","E","E","E"],["E","E","M","E","E"],["E","E","E","E","E"],["E","E","E","E","E"]], click = [3,0]
输出：[["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]
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示例 2：
输入：board = [["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]], click = [1,2]
输出：[["B","1","E","1","B"],["B","1","X","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]
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提示：
m == board.length
n == board[i].length
1 <= m, n <= 50
board[i][j] 为 'M'、'E'、'B' 或数字 '1' 到 '8' 中的一个
click.length == 2
0 <= clickr < m
0 <= clickc < n
board[clickr][clickc] 为 'M' 或 'E'
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来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/minesweeper 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
思路
本题同样是dfs，每次搜索是只需判断当前节点的周围是否有雷，如果有则填上数字然后返回，如果没有则继续搜索相邻的网格。
class Solution {
int[][] directions = {
{-1, 0}, {1, 0}, {0, 1}, {0, -1},
{1, 1}, {1, -1}, {-1, -1}, {-1, 1}
};
char[][] board = null;
int m = 0, n = 0;
public char[][] updateBoard(char[][] board, int[] click) {
this.board = board;
this.m = board.length;
this.n = board[0].length;
int x = click[0], y = click[1];
if (board[x][y] == 'M') {
board[x][y] = 'X'; return board;
}
dfs(x, y);
return board;
}
void dfs(int x, int y) {
if (board[x][y] == 'M' || board[x][y] == 'B') return;
int mines = countMine(x, y);
if (mines != 0) board[x][y] = (char)('0' + mines);
else {
board[x][y] = 'B';
for (int d = 0; d < 8; d++) {
int newX = x + directions[d][0];
int newY = y + directions[d][1];
if (0 <= newX && newX < m && 0 <= newY && newY < n)
dfs(newX, newY);
}
}
}
int countMine(int x, int y) {
int count = 0;
for (int d = 0; d < 8; d++) {
int newX = x + directions[d][0];
int newY = y + directions[d][1];
if (0 <= newX && newX < m && 0 <= newY && newY < n && board[newX][newY] == 'M')
count++;
}
return count;
}
}
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复杂度
时空复杂度应当是。
