Question

529. 扫雷游戏

English Version

题目描述

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：

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' 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

方法一：DFS

我们记 $click = (i, j)$，如果 $board[i][j]$ 等于 'M'，那么直接将 $board[i][j]$ 的值改为 'X' 即可。否则，我们需要统计 $board[i][j]$ 周围的地雷数量 $cnt$，如果 $cnt$ 不为 $0$，那么将 $board[i][j]$ 的值改为 $cnt$ 的字符串形式。否则，将 $board[i][j]$ 的值改为 'B'，并且递归地搜索处理 $board[i][j]$ 周围的未挖出的方块。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是二维数组 $board$ 的行数和列数。

class Solution:
  def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
    def dfs(i: int, j: int):
      cnt = 0
      for x in range(i - 1, i + 2):
        for y in range(j - 1, j + 2):
          if 0 <= x < m and 0 <= y < n and board[x][y] == "M":
            cnt += 1
      if cnt:
        board[i][j] = str(cnt)
      else:
        board[i][j] = "B"
        for x in range(i - 1, i + 2):
          for y in range(j - 1, j + 2):
            if 0 <= x < m and 0 <= y < n and board[x][y] == "E":
              dfs(x, y)

    m, n = len(board), len(board[0])
    i, j = click
    if board[i][j] == "M":
      board[i][j] = "X"
    else:
      dfs(i, j)
    return board

class Solution {
  private char[][] board;
  private int m;
  private int n;

  public char[][] updateBoard(char[][] board, int[] click) {
    m = board.length;
    n = board[0].length;
    this.board = board;
    int i = click[0], j = click[1];
    if (board[i][j] == 'M') {
      board[i][j] = 'X';
    } else {
      dfs(i, j);
    }
    return board;
  }

  private void dfs(int i, int j) {
    int cnt = 0;
    for (int x = i - 1; x <= i + 1; ++x) {
      for (int y = j - 1; y <= j + 1; ++y) {
        if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M') {
          ++cnt;
        }
      }
    }
    if (cnt > 0) {
      board[i][j] = (char) (cnt + '0');
    } else {
      board[i][j] = 'B';
      for (int x = i - 1; x <= i + 1; ++x) {
        for (int y = j - 1; y <= j + 1; ++y) {
          if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E') {
            dfs(x, y);
          }
        }
      }
    }
  }
}

class Solution {
public:
  vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
    int m = board.size(), n = board[0].size();
    int i = click[0], j = click[1];

    function<void(int, int)> dfs = [&](int i, int j) {
      int cnt = 0;
      for (int x = i - 1; x <= i + 1; ++x) {
        for (int y = j - 1; y <= j + 1; ++y) {
          if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M') {
            ++cnt;
          }
        }
      }
      if (cnt) {
        board[i][j] = cnt + '0';
      } else {
        board[i][j] = 'B';
        for (int x = i - 1; x <= i + 1; ++x) {
          for (int y = j - 1; y <= j + 1; ++y) {
            if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E') {
              dfs(x, y);
            }
          }
        }
      }
    };

    if (board[i][j] == 'M') {
      board[i][j] = 'X';
    } else {
      dfs(i, j);
    }
    return board;
  }
};

func updateBoard(board [][]byte, click []int) [][]byte {
  m, n := len(board), len(board[0])
  i, j := click[0], click[1]

  var dfs func(i, j int)
  dfs = func(i, j int) {
    cnt := 0
    for x := i - 1; x <= i+1; x++ {
      for y := j - 1; y <= j+1; y++ {
        if x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M' {
          cnt++
        }
      }
    }
    if cnt > 0 {
      board[i][j] = byte(cnt + '0')
      return
    }
    board[i][j] = 'B'
    for x := i - 1; x <= i+1; x++ {
      for y := j - 1; y <= j+1; y++ {
        if x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E' {
          dfs(x, y)
        }
      }
    }
  }

  if board[i][j] == 'M' {
    board[i][j] = 'X'
  } else {
    dfs(i, j)
  }
  return board
}

function updateBoard(board: string[][], click: number[]): string[][] {
  const m = board.length;
  const n = board[0].length;
  const [i, j] = click;

  const dfs = (i: number, j: number) => {
    let cnt = 0;
    for (let x = i - 1; x <= i + 1; ++x) {
      for (let y = j - 1; y <= j + 1; ++y) {
        if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] === 'M') {
          ++cnt;
        }
      }
    }
    if (cnt > 0) {
      board[i][j] = cnt.toString();
      return;
    }
    board[i][j] = 'B';
    for (let x = i - 1; x <= i + 1; ++x) {
      for (let y = j - 1; y <= j + 1; ++y) {
        if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] === 'E') {
          dfs(x, y);
        }
      }
    }
  };

  if (board[i][j] === 'M') {
    board[i][j] = 'X';
  } else {
    dfs(i, j);
  }
  return board;
}