529. Minesweeper

▶ 扫雷的扩展判定。已知棋盘上所有点的情况（雷区 'M'，已翻开空白区 'B'，未翻开空白区 'E'，数字区 '1' ~ '8'），现在给定一个点击位置（一定在空白区域），若命中雷区则将被命中的 M 改为 X，若命中空白区则将点击位置扩展为带有数字边界的安全区。

● 自己的解法，28 ms，深度优先遍历。改善了边界判定的方法，以后写类似的矩阵函数的时候可以借鉴。实际上可以在 extend 开头判定 click 是否在棋盘范围内，以后就可以强行 8 个方向搜索（见后面大佬的代码）

 class Solution
 {
 public:
     vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click)
     {
         if (board[click[]][click[]] == 'M')
         {
             board[click[]][click[]] = 'X';
             return board;
         }
         extend(board, click);
         return board;
     }
     void extend(vector<vector<char>>& board, vector<int>& click)
     {
         const int row = board.size(), col = board[].size(), rowClick = click[], colClick = click[];
         char location = ~;
         int count = ;
         vector<int> tempClick;
         // 计算当前位置的相邻情况，location从左边起 8 位分别表示 右，右上，上，左上，左，左下，下，右下 是否有相邻块
         if (colClick % col == col - )  // 右，11000001
             location &= ~;
         if (rowClick == )              // 上，01110000
             location &= ~;
         if (colClick % col == )        // 左，00011100
             location &= ~;
         if (rowClick == row - )        // 下，00000111
             location &= ~;
         // 统计周围雷数计数，从右开始，逆时针方向搜索
         if (location &  <<  && board[rowClick][colClick + ] == 'M')
             count++;
         if (location &  <<  && board[rowClick - ][colClick + ] == 'M')
             count++;
         if (location &  <<  && board[rowClick - ][colClick] == 'M')
             count++;
         if (location &  <<  && board[rowClick - ][colClick - ] == 'M')
             count++;
         if (location &  <<  && board[rowClick][colClick - ] == 'M')
             count++;
         if (location &  <<  && board[rowClick + ][colClick - ] == 'M')
             count++;
         if (location &  <<  && board[rowClick + ][colClick] == 'M')
             count++;
         if (location &  <<  && board[rowClick + ][colClick + ] == 'M')
             count++;
         if (count)// 周围有雷，本地为数字，停止搜索
         {
             board[rowClick][colClick] = count + '';
             return;
         }
         board[rowClick][colClick] = 'B';// 周围无雷，本地为安全区，继续搜索
         if (location &  <<  && board[rowClick][colClick + ] == 'E')
             extend(board, tempClick = { rowClick, colClick +  });
         if (location &  <<  && board[rowClick - ][colClick + ] == 'E')
             extend(board, tempClick = { rowClick - , colClick +  });
         if (location &  <<  && board[rowClick - ][colClick] == 'E')
             extend(board, tempClick = { rowClick - , colClick });
         if (location &  <<  && board[rowClick - ][colClick - ] == 'E')
             extend(board, tempClick = { rowClick - , colClick -  });
         if (location &  <<  && board[rowClick][colClick - ] == 'E')
             extend(board, tempClick = { rowClick, colClick -  });
         if (location &  <<  && board[rowClick + ][colClick - ] == 'E')
             extend(board, tempClick = { rowClick + , colClick -  });
         if (location &  <<  && board[rowClick + ][colClick] == 'E')
             extend(board, tempClick = { rowClick + , colClick });
         if (location &  <<  && board[rowClick + ][colClick + ] == 'E')
             extend(board, tempClick = { rowClick + , colClick +  });
         return;
     }
 };

● 大佬的代码，38 ms，深度优先遍历，与后面的广度优先遍历在格式上保持一致

 class Solution
 {
 public:
     vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click)
     {
         int m = board.size(), n = board[].size(), row = click[], col = click[];
         int count, i, j, r, c;
         vector<int>tempClick;
         if (board[row][col] == 'M')
         {
             board[row][col] = 'X';
             return board;
         }
         for (count = , i = -; i < ; i++)
         {
             for (j = -; j < ; j++)
             {
                 if (i ==  && j == )
                     continue;
                 r = row + i, c = col + j;
                 if (r <  || r >= m || c <  || c <  || c >= n)
                     continue;
                 if (board[r][c] == 'M' || board[r][c] == 'X')
                     count++;
             }
         }
         if (count)
         {
             board[row][col] = (char)(count + '');
             return board;
         }
         for (board[row][col] = 'B', i = -; i < ; i++)
         {
             for (j = -; j < ; j++)
             {
                 if (i ==  && j == )
                     continue;
                 r = row + i, c = col + j;
                 if (r <  || r >= m || c <  || c <  || c >= n)
                     continue;
                 if (board[r][c] == 'E')
                     updateBoard(board, tempClick = { r, c });
             }
         }
         return board;
     }
 };

● 大佬的广度优先遍历，31 ms，最快的解法算法与之相同，但维护一个 unordered_set<int> 用于保存已经访问过的点来防止重复访问

 class Solution
 {
 public:
     vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click)
     {
         const int m = board.size(), n = board[].size();
         queue<vector<int>> q;
         vector<int> cell;
         int row, col, count, i, j, r, c;
         for (q.push(click); !q.empty();)
         {
             cell = q.front(),q.pop();
             row = cell[], col = cell[];
             if (board[row][col] == 'M')
             {
                 board[row][col] = 'X';
                 continue;
             }
             for (count = , i = -; i < ; i++)
             {
                 for (j = -; j < ; j++)
                 {
                     if (i ==  && j == )
                         continue;
                     r = row + i, c = col + j;
                     if (r <  || r >= m || c <  || c <  || c >= n)
                         continue;
                     if (board[r][c] == 'M' || board[r][c] == 'X')
                         count++;
                 }
             }
             if (count)
             {
                 board[row][col] = (char)(count + '');
                 continue;
             }
             for (board[row][col] = 'B', i = -; i < ; i++)
             {
                 for (j = -; j < ; j++)
                 {
                     if (i ==  && j == )
                         continue;
                     r = row + i, c = col + j;
                     if (r <  || r >= m || c <  || c <  || c >= n)
                         continue;
                     if (board[r][c] == 'E')
                     {
                         q.push(vector<int>{r, c});
                         board[r][c] = 'B';
                     }
                 }
             }
         }
         return board;
     }
 };