[LeetCode] 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.

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：

解法一：

class Solution {
public:
    void solve(vectorchar> >& board) {
        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[i].size(); ++j) {
                if ((i == 0 || i == board.size() - 1 || j == 0 || j == board[i].size() - 1) && board[i][j] == 'O')
                    solveDFS(board, i, j);
            }
        }
        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[i].size(); ++j) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == '$') board[i][j] = 'O';
            }
        }
    }
    void solveDFS(vectorchar> > &board, int i, int j) {
        if (board[i][j] == 'O') {
            board[i][j] = '$';
            if (i > 0 && board[i - 1][j] == 'O') 
                solveDFS(board, i - 1, j);
            if (j < board[i].size() - 1 && board[i][j + 1] == 'O') 
                solveDFS(board, i, j + 1);
            if (i < board.size() - 1 && board[i + 1][j] == 'O') 
                solveDFS(board, i + 1, j);
            if (j > 1 && board[i][j - 1] == 'O') 
                solveDFS(board, i, j - 1);
        }
    }
};

有网友提问上面的代码中红色部分为啥是j > 1 而不是j > 0，为啥j > 0无法通过OJ的最后一个大数据集合，我开始也不知道其中奥秘，直到被另一个网友提醒在本地机子上可以通过最后一个大数据集合，于是我也写了一个程序来验证，请参见验证LeetCode Surrounded Regions 包围区域的DFS方法。发现j > 0是正确的，可以得到相同的结果。

解法二：

class Solution {
public:
    void solve(vectorchar>>& board) {
        if (board.empty() || board[0].empty()) return;
        int m = board.size(), n = board[0].size();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                    if (board[i][j] == 'O') dfs(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] == '$') board[i][j] = 'O';
            }
        }
    }
    void dfs(vectorchar>> &board, int x, int y) {
        int m = board.size(), n = board[0].size();
        vectorint>> dir{{0,-1},{-1,0},{0,1},{1,0}};
        board[x][y] = '$';
        for (int i = 0; i < dir.size(); ++i) {
            int dx = x + dir[i][0], dy = y + dir[i][1];
            if (dx >= 0 && dx < m && dy > 0 && dy < n && board[dx][dy] == 'O') {
                dfs(board, dx, dy);
            }
        }
    }
};

本文转自博客园Grandyang的博客，原文链接：包围区域[LeetCode] Surrounded Regions ，如需转载请自行联系原博主。