leetcode 529. Minesweeper 扫雷游戏 + 经典的DFS深度优先遍历

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.

输入的位置有两种可能：
1. ‘M’，说明此处是地雷，该位置置为’X’，游戏结束，返回board；
2. ‘E’，说明此处没有地雷，这种又分为：
(1) 周围8邻接范围有地雷，输出数字
(2) 周围8邻接范围没有地雷，输出’B’，翻开周围8个方块。

第一种情况可以直接输出然后游戏结束；而第二种情况需要用到DFS。
从这个方块访问周围的8个方块，如果有地雷，则count++，输出数字；如果没有地雷，输出’B’，然后需要将周围的8个方块都翻开，此处用DFS遍历这些结方块，确定这些方块的状态。

直接DFS深度优先遍历即可

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;



class Solution
{
public:
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click)
    {
        int x = click[0], y = click[1];
        if (board[x][y] == 'M')
            board[x][y] = 'X';
        else
            dfs(board, x, y);
        return board;
    }

    void dfs(vector<vector<char>>& board, int x, int y)
    {
        if (isValid(x, y, board) == false)
            return;
        else if (board[x][y] == 'E')
        {
            vector<vector<int> > surr = { { -1,0 },{ -1,-1 },{ -1,1 },{ 1,0 },{ 1,-1 },{ 1,1 },{ 0,-1 },{ 0,1 } };
            int count = 0;
            for (vector<int> i : surr)
            {
                int x1 = x + i[0], y1 = y + i[1];
                if (isValid(x1, y1, board) == true && board[x1][y1] == 'M')
                    count++;
            }

            if (count >= 1)
            {
                board[x][y] = '0' + count;
                return;
            }
            else
            {
                board[x][y] = 'B';
                for (vector<int> i : surr)
                {
                    int x1 = x + i[0], y1 = y + i[1];
                    dfs(board, x1, y1);
                }
            }
        }
    }

    bool isValid(int x, int y, vector<vector<char>>& board)
    {
        int row = board.size(), col = board[0].size();
        if (x < 0 || x >= row || y < 0 || y >= col)
            return false;
        else
            return true;
    }
};