目录

0529:扫雷游戏(★)

力扣第 529 题

题目

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''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:

  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 未挖出 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''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]'M''E''B' 或数字 '1''8' 中的一个
  • click.length == 2
  • 0 <= clickr < m
  • 0 <= clickc < n
  • board[clickr][clickc]'M''E'

相似问题:

分析

  • 如果点到地雷了,直接改为 ‘X’ 返回
  • 否则,每一步应该计算相邻地雷的个数
    • 为零就改为 ‘B’ 递归遍历所有相邻的空方块,
    • 不为零就改为数字
  • 可以用 dfs 解决,也可以用 bfs
    • 用 bfs 注意入队时就应该标记,避免重复添加

解答

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class Solution:
    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
        B,(i,j) = board,click
        m, n = len(B),len(B[0])
        if B[i][j] == 'M':
            B[i][j] = 'X'
            return B
        B[i][j] = 'B'
        Q = deque([(i,j)])
        while Q:
            i,j = Q.popleft()
            A = [(x,y) for x in range(i-1,i+2) for y in range(j-1,j+2) if 0<=x<m and 0<=y<n]
            cnt = sum(B[x][y]=='M' for x,y in A)
            if cnt:
                B[i][j] = str(cnt)
            else:
                for x,y in A:
                    if B[x][y] == 'E':
                        Q.append((x,y))
                        B[x][y] = 'B'
        return B

46 ms