Surrounded Regions 包围区域——dfs

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

这道题有点像围棋，将包住的O都变成X，但不同的是边缘的O不算被包围，跟之前那道Number of Islands 岛屿的数量很类似，都可以用DFS来解。刚开始我的思路是DFS遍历中间的O，如果没有到达边缘，都变成X，如果到达了边缘，将之前变成X的再变回来。但是这样做非常的不方便，在网上看到大家普遍的做法是扫面矩阵的四条边，如果有O，则用DFS遍历，将所有连着的O都变成另一个字符，比如，这样剩下的O都是被包围的，然后将这些O变成X，把，这样剩下的O都是被包围的，然后将这些O变成X，把变回O就行了。代码如下：

 class Solution {
 public:
     void solve(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);
             }
         }
     }
 };