[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$，这样剩下的O都是被包围的，然后将这些O变成X，把$变回O就行了。代码如下：

 

class Solution {
public:
    void solve(vector<vector<char> >& 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] == '$') board[i][j] = 'O';
            }
        }
    }
    void solveDFS(vector<vector<char> > &board, int i, int j) {
        if (board[i][j] == 'O') {
            board[i][j] = '$';
            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);
        }
    }
};

 

参考资料：

http://leetcodesolution.blogspot.com/2014/09/leetcode-surrounded-regions.html

http://fisherlei.blogspot.com/2013/03/leetcode-surrounded-regions-solution.html

http://yucoding.blogspot.com/2013/08/leetcode-question-131-surrounded-regions.html

 

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