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 = click
        if board[r][c] == 'M':
            board[r][c] = 'X'
            return board
        
        rows = len(board)
        cols = len(board[0])

        visited = [[False] * cols for _ in range(rows)]
        visited[r][c] = True
        directions = [(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 = 0
            for dr, dc in directions:
                nr = r + dr
                nc = c + dc
                if 0 <= nr < rows and 0 <= nc < cols:
                    if board[nr][nc] == 'M':
                        count += 1
            
            if count > 0:
                board[r][c] = str(count)
            else:
                board[r][c] = 'B'
                for dr, dc in directions:
                    nr = r + dr
                    nc = c + dc
                    if 0 <= nr < rows and 0 <= nc < cols and not visited[nr][nc] and board[nr][nc] == 'E':
                        q.append((nr, nc))
                        visited[nr][nc] = True
        return 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： $O(MN)$ M和N分别为行数和列数
• 空间复杂度： $O(MN)$ M和N分别为行数和列数

DFS

python实现

class Solution:
    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
        r, c = click
        if board[r][c] == 'M':
            board[r][c] = 'X'
            return board
        
        rows = len(board)
        cols = len(board[0])

        directions = [(1, 0), (-1, 0), (0, 1), (0, -1), (1, 1), (1, -1), (-1, 1), (-1, -1)]
        def dfs(r, c):
            count = 0
            for dx, dy in directions:
                nr = r + dx
                nc = c + dy
                if 0 <= nr < rows and 0 <= nc < cols:
                    if board[nr][nc] == 'M':
                        count += 1
            
            if count > 0:
                board[r][c] = str(count)
            else:
                board[r][c] = 'B'
                for dx, dy in directions:
                    nr = r + dx
                    nc = c + dy
                    if 0 <= nr < rows and 0 <= nc < cols and board[nr][nc] == 'E':
                        dfs(nr, nc)
        dfs(r, c)
        return board

c++实现

class Solution {
    vector<pair<int, int>> directions = {
        {1, 0}, {-1, 0}, {0, 1}, {0, -1}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};

public:
    void dfs(int r, int c, vector<vector<char>>& board, int rows, int cols) {
        int count = 0;
        for(auto tmp: directions) {
            int nr = r + tmp.first;
            int nc = c + tmp.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 tmp: directions) {
                int nr = r + tmp.first;
                int nc = c + tmp.second;
                if (0 <= nr && nr < rows && 0 <= nc && nc < cols && board[nr][nc] == 'E') {
                    dfs(nr, nc, board, rows, cols);
            }
        }
    }
    }

    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();
        dfs(r, c, board, rows, cols);
        return board;
    }
};

复杂度分析

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

参考

• https://leetcode.cn/problems/minesweeper/solution/sao-lei-you-xi-by-leetcode-solution/