leetcode--21. 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 .
给定一个包含“X”和“O”的2D板，捕获所有被“X”包围的区域。捕获的意思就是说通过把被包围的区域中 'O' 替换成 'X' 来实现的。
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

思路：
四条边上的"O"和与之相连的"O"不替换

public class Solution {
    public void solve(char[][] board) {
        if (board.length <= 0)
            return;
        int row = board.length;
        int col = board[0].length;
        for (int i = 0; i != row; i++) {
            DFS(board, i, col - 1);
            DFS(board, i, 0);
        }
        for (int i = 0; i != col; i++) {
            DFS(board, 0, i);
            DFS(board, row - 1, i);
        }

        for (int i = 0; i != row; i++)
            for (int j = 0; j != col; j++) {
                if (board[i][j] == 'O')
                    board[i][j] = 'X';
                if (board[i][j] == '*')
                    board[i][j] = 'O';
            }
    }

    public void DFS(char[][] board, int row, int col) {
        if (row < 0 || row > (board.length - 1) || col < 0 || col > (board[0].length - 1))
            return;
        if (board[row][col] == 'O') {
            board[row][col] = '*';
            DFS(board, row - 1, col);
            DFS(board, row + 1, col);
            DFS(board, row, col - 1);
            DFS(board, row, col + 1);
        }
    }
}