[Leetcode] 529. Minesweeper 解题报告

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本文详细解析了一款基于DFS或BFS实现的扫雷游戏算法。通过递归地揭示空格及其相邻区域，实现了游戏板状态的更新。适用于游戏开发及算法学习。

题目：

Let's play the minesweeper game (Wikipedia, online game)!

You are given a 2D char matrix representing the game board. 'M' represents an unrevealed mine, 'E' represents an unrevealed empty square, 'B' represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, digit ('1' to '8') represents how many mines are adjacent to this revealed square, and finally 'X' represents a revealed mine.

Now given the next click position (row and column indices) 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 - change it to 'X'.
If an empty square ('E') with no adjacent mines is revealed, then change it to 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:

Input: 

[['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']]

Explanation:

Example 2:

Input: 

[['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']]

Explanation:

Note:

The range of the input matrix's height and width is [1,50].
The click position will only be an unrevealed square ('M' or 'E'), which also means the input board contains at least one clickable square.
The input board won't be a stage when game is over (some mines have been revealed).
For simplicity, not mentioned rules should be ignored in this problem. For example, you don't need to reveal all the unrevealed mines when the game is over, consider any cases that you will win the game or flag any squares.

思路：

一道比较正常的DFS或者BFS题目。个人比较喜欢DFS，因为代码更加简洁。需要注意的是，对于一个格子，我们需要从八个方向对它进行深度搜索。

代码：

class Solution {
public:
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        if (board.size() == 0 || board[0].size() == 0) {
            return board;
        }
        int row_num = board.size(), col_num = board[0].size(), row = click[0], col = click[1];
        if (board[row][col] == 'M') {
            board[row][col] = 'X';
        }
        else {
            vector<vector<bool>> visited(row_num, vector<bool>(col_num, false));
            DFS(board, visited, row, col);
        }
        return board;
    }
private:
    void DFS(vector<vector<char>> &board, vector<vector<bool>> &visited, int row, int col) {
        if (row < 0 || row >= board.size() || col < 0 || col >= board[0].size()
            || visited[row][col] || isdigit(board[row][col])) {
            return;
        }
        int mine_num = mineNumber(board, row, col);
        board[row][col] = mine_num == 0 ? 'B' : '0' + mine_num;
        visited[row][col] = true;
        if (mine_num == 0) {
            DFS(board, visited, row - 1, col - 1);  // up-left
            DFS(board, visited, row - 1, col);      // up
            DFS(board, visited, row - 1, col + 1);  // up_right
            DFS(board, visited, row, col - 1);      // left
            DFS(board, visited, row, col + 1);      // right
            DFS(board, visited, row + 1, col - 1);  // down_left
            DFS(board, visited, row + 1, col);      // down
            DFS(board, visited, row + 1, col + 1);  // down_right
        }
    }
    int mineNumber(vector<vector<char>> &board, int row, int col) {
        int ret = 0;
        for (int i = -1; i <= 1; ++i) {
            for (int j = -1; j <= 1; ++j) {
                if (i != 0 || j != 0) {
                    int r = row + i, c = col + j;
                    if (r >= 0 && r < board.size() && c >= 0 && c < board[0].size() && board[r][c] == 'M') {
                        ++ret;
                    }
                }
            }
        }
        return ret;
    }
};