leetcode: 529. 扫雷游戏

529. 扫雷游戏

来源：力扣（LeetCode）

链接: https://leetcode.cn/problems/number-of-islands/

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"]]

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"]]

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'

解法

总的思想是进行遍历，从最开始点击开始，然后看其什么，然后看其八个方向，这里是八个方向，不是四个方向遍历，然后统计地雷的个数，如果遍历完毕后地雷数不为0，然后改变该处的位置为对应的count个数的字符，如果为0，则往八个方向

BFS：用队列进行存储该陆地，并进行八个方向的遍历，如果后续有不为地雷的位置，进入队列，直到队列为空
DFS：用递归的方式进行八个方向的遍历

代码实现

BFS

python实现

class Solution:def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:r, c = clickif board[r][c] == 'M':board[r][c] = 'X'return boardrows = len(board)cols = len(board[0])visited = [[False] * cols for _ in range(rows)]visited[r][c] = Truedirections = [(1, 0), (-1, 0), (0, 1), (0, -1), (1, 1), (1, -1), (-1, -1), (-1, 1)]q = [(r, c)]while q:r, c = q.pop(0)count = 0for dr, dc in directions:nr = r + drnc = c + dcif 0 <= nr < rows and 0 <= nc < cols:if board[nr][nc] == 'M':count += 1if count > 0:board[r][c] = str(count)else:board[r][c] = 'B'for dr, dc in directions:nr = r + drnc = c + dcif 0 <= nr < rows and 0 <= nc < cols and not visited[nr][nc] and board[nr][nc] == 'E':q.append((nr, nc))visited[nr][nc] = Truereturn board

c++实现

class Solution {public:vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {int r = click[0], c = click[1];if (board[r][c] == 'M') {board[r][c] = 'X';return board;}int rows = board.size();int cols = board[0].size();vector<pair<int, int>> directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};vector<vector<int>> visited(rows, vector<int>(board[0].size(), 0));queue<pair<int, int>> q;q.push({r, c});while (q.size() != 0) {int count = 0;auto tmp = q.front();q.pop();int r = tmp.first;int c = tmp.second;for (auto dr: directions) {int nr = r + dr.first;int nc = c + dr.second;if (0 <= nr && nr < rows && 0 <= nc && nc < cols) {if (board[nr][nc] == 'M')count++;}}if (count > 0) board[r][c] = count + '0';else {board[r][c] = 'B';for (auto dr: directions) {int nr = r + dr.first;int nc = c + dr.second;if (0 <= nr  && nr < rows && 0 <= nc && nc < cols && !visited[nr][nc] && board[nr][nc] == 'E') {q.push({nr, nc});visited[nr][nc] = true;}}}}return board;}
};

复杂度分析

时间复杂度： O(MN)O(MN)O(MN) M和N分别为行数和列数
空间复杂度： O(MN)O(MN)O(MN) M和N分别为行数和列数

DFS