130. Surrounded Regions

该题是要将被X包围的O区域用X覆盖，处于边界的O区域不算是被X包围。

主要解决思路是遍历边框的的元素，如果是O，那么从该区域开始，进行深度遍历，将是O的区域改成Y，表示这个区域是没有被包围的。完成边框遍历后，剩下的O区域都是被X包围的了。对所有元素进行遍历，将O改成X， 将Y改回O即可。

代码如下：

class Solution {
    public void solve(char[][] board) {
        if( board.length == 0 || board == null ){
            return;
        }
        
        int m = board.length, n = board[0].length;
        
        for(int i = 0, j = 0 ; i < m; i++){
            if(board[i][j] == 'O'){
                occupy(board, i, j);
            }
        }
        
        for(int i = 0, j = n - 1 ; i < m; i++){
            if(board[i][j] == 'O'){
                occupy(board, i, j);
            }           
        }
        
        for(int i = 0, j = 0; j < n ; j++){
            if(board[i][j] == 'O'){
                occupy(board, i, j);
            }            
        }
        
        for(int i = m - 1, j = 0 ; j < n; j++){
            if(board[i][j] == 'O'){
                occupy(board, i, j);
            }            
        }
        
        for(int i = 0 ; i < m ; i++){
            for(int j = 0; j < n ; j++){
                if(board[i][j] == 'O'){
                    board[i][j] = 'X';
                }
                
                if(board[i][j] == 'Y'){
                    board[i][j] = 'O';
                }
            }
        }
        
    }
    
    private void occupy(char[][] board, int i, int j){
        board[i][j] = 'Y';
        
        int m = board.length, n = board[0].length;
        if( i + 1 < m && board[i+1][j] == 'O'){
            occupy(board, i + 1, j);
        }
        
        if( i - 1 >= 0 && board[i-1][j] == 'O' ){
            occupy(board, i - 1, j);
        }
        
        if( j + 1 < n && board[i][j+1] == 'O'){
            occupy(board, i , j + 1);
        }
        
        if( j - 1 >= 0 && board[i][j-1] == 'O'){
            occupy(board, i, j - 1);
        }
    }
}

END