LeetCode_Surrounded Regions

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region .
For example,

X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X

　　solution： 思想是mark 出所有外围的O，里面的不管。

1. First scan the four edges of the board, if you meet an 'O', call a recursive mark function to mark that region to something else (for example, '+');
2. scan all the board, if you meet an 'O', flip it to 'X';
3. scan again the board, if you meet an '+', flip it to 'O';

   class Solution {
   public:
        void BFS(vector<vector<char>> &board, int i , int j){
   
           if(board[i][j] == 'X' || board[i][j] == '#') return;
           board[i][j] = '#';
   
           if( i - >=  ) BFS(board, i-, j );
           if( i +  <  rows  ) BFS(board, i+, j );
           if( j - >=  ) BFS(board, i, j- );
           if( j + < columns ) BFS(board, i, j+ );
       }
       void solve(vector<vector<char>> &board) {
           // Start typing your C/C++ solution below
           // DO NOT write int main() function
           rows = board.size();
           if( rows == ) return ;
           columns = board[].size();
           if(columns == ) return ;
   
           for( int i = ; i < columns -  ; i++)
           {
               BFS(board, ,i);//up
               BFS(board, rows -  , i );//down
           }
           for( int i = ; i< rows; i++)
           {
               BFS(board, i,  ); //left
               BFS(board, i, columns - ); //right
           }
           for(int i = ; i< rows; i++)
               for(int j = ; j < columns ; j++)
               {
                 if(board[i][j] == '#')
                    board[i][j] = 'O';
                 else if (board[i][j] == 'O')
                   board[i][j] = 'X';
               }
       }
   private :
       int rows;
       int columns;
   };