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529. 扫雷游戏

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：

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：

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上次编辑到这里，代码来自缓存 点击恢复默认模板
class Solution { public: vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) { } };

python3 解法, 执行用时: 64 ms, 内存消耗: 15.4 MB, 提交时间: 2022-11-28 14:48:18 

class Solution:
    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
        i, j = click
        row, col = len(board), len(board[0])
        if board[i][j] == "M":
            board[i][j] = "X"
            return board

        # 计算空白快周围的***
        def cal(i, j):
            res = 0
            for x in [1, -1, 0]:
                for y in [1, -1, 0]:
                    if x == 0 and y == 0: continue
                    if 0 <= i + x < row and 0 <= j + y < col and board[i + x][j + y] == "M": res += 1
            return res

        def bfs(i, j):
            queue = collections.deque([[i, j]])
            while queue:
                i, j = queue.pop()
                num = cal(i, j)
                if num > 0:
                    board[i][j] = str(num)
                    continue
                board[i][j] = "B"
                for x in [1, -1, 0]:
                    for y in [1, -1, 0]:
                        if x == 0 and y == 0: continue
                        nxt_i, nxt_j = i + x, j + y
                        if nxt_i < 0 or nxt_i >= row or nxt_j < 0 or nxt_j >= col: continue 
                        if board[nxt_i][nxt_j] == "E":
                            queue.appendleft([nxt_i, nxt_j])
                            board[nxt_i][nxt_j] = "B"

        bfs(i, j)
        return board

python3 解法, 执行用时: 48 ms, 内存消耗: 19.1 MB, 提交时间: 2022-11-28 14:48:05 

class Solution:
    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
        i, j = click
        row, col = len(board), len(board[0])
        if board[i][j] == "M":
            board[i][j] = "X"
            return board

        # 计算空白快周围的***
        def cal(i, j):
            res = 0
            for x in [1, -1, 0]:
                for y in [1, -1, 0]:
                    if x == 0 and y == 0: continue
                    if 0 <= i + x < row and 0 <= j + y < col and board[i + x][j + y] == "M": res += 1
            return res

        def dfs(i, j):
            num = cal(i, j)
            if num > 0:
                board[i][j] = str(num)
                return
            board[i][j] = "B"
            for x in [1, -1, 0]:
                for y in [1, -1, 0]:
                    if x == 0 and y == 0: continue
                    nxt_i, nxt_j = i + x, j + y
                    if 0 <= nxt_i < row and 0 <= nxt_j < col and board[nxt_i][nxt_j] == "E": dfs(nxt_i, nxt_j)

        dfs(i, j)
        return board

golang 解法, 执行用时: 24 ms, 内存消耗: 6.5 MB, 提交时间: 2022-11-28 14:47:15 

var dirX = []int{0, 1, 0, -1, 1, 1, -1, -1}
var dirY = []int{1, 0, -1, 0, 1, -1, 1, -1}

func updateBoard(board [][]byte, click []int) [][]byte {
    x, y := click[0], click[1]
    if board[x][y] == 'M' {
        board[x][y] = 'X'
    } else {
        bfs(board, x, y)
    }
    return board
}

func bfs(board [][]byte, sx, sy int) {
    queue := [][]int{}
    vis := make([][]bool, len(board))
    for i := 0; i < len(vis); i++ {
        vis[i] = make([]bool, len(board[0]))
    }
    queue = append(queue, []int{sx, sy})
    vis[sx][sy] = true
    for i := 0; i < len(queue); i++ {
        cnt, x, y := 0, queue[i][0], queue[i][1]
        for i := 0; i < 8; i++ {
        tx, ty := x + dirX[i], y + dirY[i]
            if tx < 0 || tx >= len(board) || ty < 0 || ty >= len(board[0]) {
                continue
            }
            // 不用判断 M，因为如果有 M 的话游戏已经结束了
            if board[tx][ty] == 'M' {
                cnt++
            }
        }
        if cnt > 0 {
            board[x][y] = byte(cnt + '0')
        } else {
            board[x][y] = 'B'
            for i := 0; i < 8; i++ {
                tx, ty := x + dirX[i], y + dirY[i]
                // 这里不需要在存在 B 的时候继续扩展，因为 B 之前被点击的时候已经被扩展过了
                if tx < 0 || tx >= len(board) || ty < 0 || ty >= len(board[0]) || board[tx][ty] != 'E' || vis[tx][ty] {
                    continue
                }
                queue = append(queue, []int{tx, ty})
                vis[tx][ty] = true
            }
        }
    }
}

golang 解法, 执行用时: 24 ms, 内存消耗: 6.9 MB, 提交时间: 2022-11-28 14:47:01 

var dirX = []int{0, 1, 0, -1, 1, 1, -1, -1}
var dirY = []int{1, 0, -1, 0, 1, -1, 1, -1}

func updateBoard(board [][]byte, click []int) [][]byte {
    x, y := click[0], click[1]
    if board[x][y] == 'M' {
        board[x][y] = 'X'
    } else {
        dfs(board, x, y)
    }
    return board
}

func dfs(board [][]byte, x, y int) {
    cnt := 0
    for i := 0; i < 8; i++ {
        tx, ty := x + dirX[i], y + dirY[i]
        if tx < 0 || tx >= len(board) || ty < 0 || ty >= len(board[0]) {
            continue
        }
        // 不用判断 M，因为如果有 M 的话游戏已经结束了
        if board[tx][ty] == 'M' {
            cnt++
        }
    }
    if cnt > 0 {
        board[x][y] = byte(cnt + '0')
    } else {
        board[x][y] = 'B'
        for i := 0; i < 8; i++ {
            tx, ty := x + dirX[i], y + dirY[i]
            // 这里不需要在存在 B 的时候继续扩展，因为 B 之前被点击的时候已经被扩展过了
            if tx < 0 || tx >= len(board) || ty < 0 || ty >= len(board[0]) || board[tx][ty] != 'E' {
                continue
            }
            dfs(board, tx, ty)
        }
    }
}

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