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Kotlin-based LeetCode algorithm problem solutions, regularly updated
529\. Minesweeper
Medium
Let's play the minesweeper game ([Wikipedia](https://en.wikipedia.org/wiki/Minesweeper_(video_game)), [online game](http://minesweeperonline.com))!
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_:
1. If a mine `'M'` is revealed, then the game is over. You should change it to `'X'`.
2. 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.
3. 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.
4. Return the board when no more squares will be revealed.
**Example 1:**
![](https://assets.leetcode.com/uploads/2018/10/12/minesweeper_example_1.png)
**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://assets.leetcode.com/uploads/2018/10/12/minesweeper_example_2.png)
**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'`.