LeetCode 529. Minesweeper

529. Minesweeper

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 unrevealedempty 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 adjacentunrevealed 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:

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:

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.

题意：

类似于扫雷游戏，给你一个地图和一系列点击位置，然后更新地图状态

M代表地雷，点击到的话更新为X

E代表未探明的安全区，点击到的话，有下面两种情况：

1）如果周围8个方格没有地雷，则更新为B，然后继续去探查周围的方格

2）如果周围8个方格有地雷，则更新为地雷的个数

代码：

class Solution {
public:
    vector<vector<int>>vis;//用来标记访问过的位置
    //8个方向的偏移量数组
    int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
    void dfs(vector<vector<char>>& board, int x, int y, int m, int n){
        if(board[x][y] == 'M'){//如果踩到地雷，标记为X
            board[x][y] = 'X';
            return ;
        }
        vis[x][y] = 1;//将（x, y）标记为访问过
        int sum = 0;
        for(int i = 0; i < 8; i++){//查找（x, y）周围方格里有几个地雷
            int dx = x + dir[i][0];
            int dy = y + dir[i][1];
            if(dx >= 0 && dx < m && dy >= 0 && dy < n && board[dx][dy] == 'M')
                sum++;
        }
        if(sum > 0){
            board[x][y] = '0' + sum;//（x, y）周围有地雷的话，就将该位置标记为地雷个数
        }else{
            board[x][y] = 'B';//否则将该位置标记为安全区，然后继续探查周围的方格
            for(int i = 0; i < 8; i++){
                int dx = x + dir[i][0];
                int dy = y + dir[i][1];
                if(dx >= 0 && dx < m && dy >= 0 && dy < n && vis[dx][dy] == 0)
                    dfs(board, dx, dy, m, n);
            }
        }
    }
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        int m = board.size();
        int n = board[0].size();
        vis.assign(m,vector<int>(n,0));//初始化标记数组
        int x, y;
        for(int i = 0; i < click.size(); i+= 2){
            x = click[i];
            y = click[i+1];
            dfs(board, x, y, m, n);
        }
        return board;
    }
};