LeetCode_Surrounded Regions

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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 -1 >= 0 ) BFS(board, i-1, j );
           if( i + 1 <  rows  ) BFS(board, i+1, j ); 
           if( j -1 >= 0 ) BFS(board, i, j-1 );
           if( j +1 < columns ) BFS(board, i, j+1 );
       }
       void solve(vector<vector<char>> &board) {
           // Start typing your C/C++ solution below
           // DO NOT write int main() function
           rows = board.size();
           if( rows == 0) return ;
           columns = board[0].size();
           if(columns == 0) return ;
           
           for( int i = 1; i < columns - 1 ; i++)
           {
               BFS(board, 0,i);//up
               BFS(board, rows - 1 , i );//down
           }
           for( int i = 0; i< rows; i++)
           {
               BFS(board, i, 0 ); //left
               BFS(board, i, columns - 1); //right
           }
           for(int i = 0; i< rows; i++)
               for(int j = 0; 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;
   };