Description#
Let's play the minesweeper game (Wikipedia, online game)!
You are given an m x n
char matrix board
representing the game board where:
'M'
represents an unrevealed mine,'E'
represents an unrevealed empty square,'B'
represents a revealed blank square that has no adjacent mines (i.e., above, below, left, right, and all 4 diagonals),- digit (
'1'
to '8'
) represents how many mines are adjacent to this revealed square, and 'X'
represents a revealed mine.
You are also given an integer array click
where click = [clickr, clickc]
represents the next click position among all the unrevealed squares ('M'
or 'E'
).
Return the board after revealing this position according to the following rules:
- If a mine
'M'
is revealed, then the game is over. You should change it to 'X'
. - If an empty square
'E'
with no adjacent mines is revealed, then change it to a revealed blank 'B'
and all of its adjacent unrevealed squares should be revealed recursively. - If an empty square
'E'
with at least one adjacent mine is revealed, then change it to a digit ('1'
to '8'
) representing the number of adjacent mines. - Return the board when no more squares will be revealed.
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/0529.Minesweeper/images/untitled.jpeg)
Input: board = [["E","E","E","E","E"],["E","E","M","E","E"],["E","E","E","E","E"],["E","E","E","E","E"]], click = [3,0]
Output: [["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/0529.Minesweeper/images/untitled-2.jpeg)
Input: board = [["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]], click = [1,2]
Output: [["B","1","E","1","B"],["B","1","X","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]
Constraints:
m == board.length
n == board[i].length
1 <= m, n <= 50
board[i][j]
is either 'M'
, 'E'
, 'B'
, or a digit from '1'
to '8'
.click.length == 2
0 <= clickr < m
0 <= clickc < n
board[clickr][clickc]
is either 'M'
or 'E'
.
Solutions#
Solution 1#
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| 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
|
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| 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);
}
}
}
}
}
}
|
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| 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;
}
};
|
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| 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
}
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| 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;
}
|