[LintCode] Surrounded Regions 包围区域

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.
Example

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

After capture all regions surrounded by 'X', the board should be:

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

LeetCode上的原题，请参见我之前的博客Surrounded Regions。

解法一：

class Solution {
public:
    /**
     * @param board a 2D board containing 'X' and 'O'
     * @return void
     */
    void surroundedRegions(vector<vector<char>>& board) {
        if (board.empty() || board[].empty()) return;
        int m = board.size(), n = board[].size();
        for (int i = ; i < m; ++i) {
            for (int j = ; j < n; ++j) {
                if (i ==  || i == m -  || j ==  || j == n - ) {
                    if (board[i][j] == 'O') dfs(board, i , j);
                }
            }
        }
        for (int i = ; i < m; ++i) {
            for (int j = ; j < n; ++j) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == '$') board[i][j] = 'O';
            }
        }
    }
    void dfs(vector<vector<char>> &board, int x, int y) {
        int m = board.size(), n = board[].size();
        vector<vector<int>> dir{{,-},{-,},{,},{,}};
        board[x][y] = '$';
        for (int i = ; i < dir.size(); ++i) {
            int dx = x + dir[i][], dy = y + dir[i][];
            if (dx >=  && dx < m && dy >=  && dy < n && board[dx][dy] == 'O') {
                dfs(board, dx, dy);
            }
        }
    }
};

解法二：

class Solution {
public:
    /**
     * @param board a 2D board containing 'X' and 'O'
     * @return void
     */
    void surroundedRegions(vector<vector<char>>& board) {
        if (board.empty() || board[].empty()) return;
        int m = board.size(), n = board[].size();
        for (int i = ; i < m; ++i) {
            for (int j = ; j < n; ++j) {
                if (i ==  || i == m -  || j ==  || j == n - ) {
                    if (board[i][j] == 'O') dfs(board, i , j);
                }
            }
        }
        for (int i = ; i < m; ++i) {
            for (int j = ; j < n; ++j) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == '$') board[i][j] = 'O';
            }
        }
    }
    void dfs(vector<vector<char>> &board, int i, int j) {
        int m = board.size(), n = board[].size();
        if (i <  || i >= m || j <  || j >= n || board[i][j] != 'O') return;
        board[i][j] = '$';
        dfs(board, i + , j);
        dfs(board, i - , j);
        dfs(board, i, j + );
        dfs(board, i, j - );
    }
};