1189. 扫雷游戏

1189. 扫雷游戏

中文English

让我们一起来玩扫雷游戏！

给定一个代表游戏板的二维字符矩阵。 '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。

现在给出在所有未挖出的方块中（'M'或者'E'）的下一个点击位置（行和列索引），根据以下规则，返回相应位置被点击后对应的面板：

1. 如果一个地雷（'M'）被挖出，游戏就结束了- 把它改为 'X'。
2. 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方块都应该被递归地揭露。
3. 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。
4. 如果在此次点击中，若无更多方块可被揭露，则返回面板。

样例

样例 1:

输入: board = ["EEEEE","EEMEE","EEEEE","EEEEE"], Click : [3,0]
输出: ["B1E1B","B1M1B","B111B","BBBBB"]

样例 2:

输入: board = ["B1E1B","B1M1B", "B111B","BBBBB"], Click : [1,2]
输出: ["B1E1B","B1X1B","B111B","BBBBB"]

注意事项

1.输入矩阵的宽和高的范围为 [1,50]。
2.点击的位置只能是未被挖出的方块 ('M' 或者 'E')，这也意味着面板至少包含一个可点击的方块。
3.输入面板不会是游戏结束的状态（即有地雷已被挖出）。
4.简单起见，未提及的规则在这个问题中可被忽略。例如，当游戏结束时你不需要挖出所有地雷，考虑所有你可能赢得游戏或标记方块的情况。

BFS写法

class Solution:
    """
    @param board: a board
    @param click: the position
    @return: the new board
    """
    def updateBoard(self, board, click):
        # Write your code here
        #bfs写法，如果是数字或者是M的话，则说明是不能再挖下去。如果是B的话，则可以加入队列继续挖出
        queue = [click]
        m, n = len(board), len(board[0])
        result = self.bfs(board, queue, m, n)

        return result

    def bfs(self, board, queue, m, n):

        while queue:
            array = queue.pop()
            x, y = array[0], array[1]
            
            #否则的话
            if board[x][y] == 'M':
                board[x] = board[x][: y] + 'X' + board[x][y + 1: ]
                continue
            
            #分情况考虑
            if board[x][y] == 'E':
                res = self.isSatified(board, x, y, m, n)
                if res:
                    board[x] = board[x][: y] + res + board[x][y + 1: ]
                else:
                    board[x] = board[x][: y] + 'B' + board[x][y + 1: ]
                
                    #只要是E就需要appeend
                    directions = [[1, 0], [-1, 0], [0, 1], [0, -1], [1, 1], [1, -1], [-1, 1], [-1, -1]]
                    for direction in directions:
                        new_x, new_y = x + direction[0], y + direction[1]
                        if new_x < 0 or new_x > m - 1 or new_y < 0 or new_y > n - 1:
                            continue 

                        if board[new_x][new_y] == 'E':
                            queue.append([new_x, new_y])

        return board

    #判断四周是否是雷
    def isSatified(self, board, x, y, m, n):
        directions = [[1, 0], [-1, 0], [0, 1], [0, -1], [1, 1], [1, -1], [-1, 1], [-1, -1]]

        count = 0
        for direction in directions:
            x1, y1 = direction[0] + x, direction[1] + y

            if x1 < 0 or x1 > m - 1 or y1 < 0 or y1 > n - 1:
                continue

            #判断是否是雷
            if board[x1][y1] == 'M':
                count += 1
        
        return str(count) if count != 0 else False

 

DFS写法

#DFS写法
class Solution:
    """
    @param board: a board
    @param click: the position
    @return: the new board
    """
    def updateBoard(self, board, click):
        # Write your code here
        directions = [[1, 0], [-1, 0], [0, 1], [0, -1], [1, 1], [1, -1], [-1, 1], [-1, -1], [0, 0]]
        m, n = len(board), len(board[0])
        self.dfs(board, m, n, click[0], click[1], directions)

        return board

    def dfs(self, board, m, n, x, y, directions):
        #如果是M的话，直接返回即可
        if board[x][y] == 'M': 
            board[x] = board[x][: y] + 'X' + board[x][y + 1: ]
            return 
        #如果是E的话, 需要判断周围存在多少个雷，如果是E则继续递归，否则continue
        for direction in directions:
            new_x, new_y = x + direction[0], y + direction[1]
            
            if 0 <= new_x < m and 0 <= new_y < n and board[new_x][new_y] == 'E':
                res = self.getAround(board, m, n, x, y, directions)
                if res:
                    board[x] = board[x][: y] + res + board[x][y + 1: ]
                    continue
                board[x] = board[x][: y] + 'B' + board[x][y + 1: ]
                self.dfs(board, m, n, new_x, new_y, directions)
        
    def getAround(self, board, m, n, x, y, directions):
        count = 0

        for direction in directions:
            new_x, new_y = direction[0] + x, direction[1] + y

            if 0 <= new_x < m and 0 <= new_y < n and board[new_x][new_y] == 'M':
                count += 1 
        
        return str(count)  if count else False

 

DFS(稍作调整)

#DFS写法
class Solution:
    """
    @param board: a board
    @param click: the position
    @return: the new board
    """
    def updateBoard(self, board, click):
        # Write your code here
        directions = [[1, 0], [-1, 0], [0, 1], [0, -1], [1, 1], [1, -1], [-1, 1], [-1, -1]]
        m, n = len(board), len(board[0])
        self.dfs(board, m, n, click[0], click[1], directions)

        return board

    def dfs(self, board, m, n, x, y, directions):
        #如果是M的话，直接返回即可
        if board[x][y] == 'M': 
            board[x] = board[x][: y] + 'X' + board[x][y + 1: ]
            return 

        res = self.getAround(board, m, n, x, y, directions)
        if res:
            board[x] = board[x][: y] + res + board[x][y + 1: ]
        else:
            board[x] = board[x][: y] + 'B' + board[x][y + 1: ]
                
            #如果不是数字的话，则说明可以继续四周递归下去, res是计算四周雷的个数
            for direction in directions:
                new_x, new_y = x + direction[0], y + direction[1]
                
                if 0 <= new_x < m and 0 <= new_y < n and board[new_x][new_y] == 'E':
                    self.dfs(board, m, n, new_x, new_y, directions)
        
    def getAround(self, board, m, n, x, y, directions):
        count = 0

        for direction in directions:
            new_x, new_y = direction[0] + x, direction[1] + y

            if 0 <= new_x < m and 0 <= new_y < n and board[new_x][new_y] == 'M':
                count += 1 
        
        return str(count) if count else False