LeetCode 529. Minesweeper 解题报告

LeetCode 529. Minesweeper 解题报告

题目描述

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:

1. If a mine (‘M’) is revealed, then the game is over - change it to ‘X’.
2. 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.
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.

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示例

Example 1:

Example 2:

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注意事项

1. The range of the input matrix’s height and width is [1,50].
2. The click position will only be an unrevealed square (‘M’ or ‘E’), which also means the input board contains at least one clickable square.
3. The input board won’t be a stage when game is over (some mines have been revealed).
4. 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.

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解题思路

我的思路：

这道题的题目很长，其实表达的意思就是给你一个地图和一系列的点击位置，需要根据点击的情况对地图进行更新。地图中M表示未探明的地雷，E表示未探明的安全区，B表示已探明且周围8个格子均没有地雷的安全区，X表示点击中了地雷，数字表示该格子周围的地雷数。更新规则是点击M则将该位置标志位X，点击E时则存在两种情况，一是将该格子更新为周围8个格子中的地雷总数，二是如果周围没有地雷则为B，并继续探测周围的格子。
看完题目就知道这是一道题迷宫地图类的题目，关键的点是点击E时周围没有地雷就需要继续探测，这里需要做的就是通过深度优先搜索（Depth-First-Search）实现持续探测。
具体实现是中规中矩的dfs，使用与地图同样大小的数组存储访问情况，避免无限递归，并使用数组存储8个邻格的坐标偏移值，每次处理某一个坐标的格子，通过循环来计算该格子周围的8个邻格是否在地图范围内，如果在地图范围内且该邻格没有访问过就以该邻格为中心进入下一次递归搜索。见下面实现代码。

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代码

我的代码

class Solution {
public:
    vector<vector<bool>> visited;
    int n,m;

    void dfs(vector<vector<char>>& board, int x, int y) {
        if (board[x][y] == 'M') {
            board[x][y] = 'X';
            return ;
        }

        if (board[x][y] != 'E')
            return ;

        int offsetX[8] = {-1, -1, -1,  0, 0,  1, 1, 1};
        int offsetY[8] = {-1,  0,  1, -1, 1, -1, 0, 1};
        int mines = 0;

        visited[x][y] = true;

        for (int i = 0; i < 8; i++) {
            int row = x + offsetX[i];
            int col = y + offsetY[i];

            if (row < 0 || row >= n || col < 0 || col >= m)
                continue;
            if (board[row][col] == 'M')
                mines++;
        }

        if (mines > 0) {
            board[x][y] = '0' + mines;
        } else {
            board[x][y] = 'B';

            for (int i = 0; i < 8; i++) {
                int row = x + offsetX[i];
                int col = y + offsetY[i];

                if (row < 0 || row >= n || col < 0 || col >= m || visited[row][col])
                    continue;
                dfs(board, row, col);
            }
        }
    }

    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        int x, y;

        n = board.size();
        m = board[0].size();

        visited.assign(n, vector<bool>(m, false));

        for (int i = 0; i < click.size(); i += 2) {
            x = click[i];
            y = click[i + 1];
            dfs(board, x, y);
        }

        return board;
    }
};

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总结

这道题是使用深度优先搜索，实现方式是递归，每次递归中只处理一个位置，并在遇到符合规则的邻格时进入下一层递归，需要注意的就是要按照地图中字符的更新规则进行更新。
这周暂时先完成一道DFS，下周会做BFS的题目，要努力加油，把算法给提升上去~