LeetCode[BFS]: 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.

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

参考：https://oj.leetcode.com/discuss/9942/my-bfs-solution-c-28ms

采取逆向思维，不是找到所有被’X’包围的’O’，而是将所有未被包围的’O’先标记为’#’，然后将剩余的’O’标记为’X’，将’#’标记为’O’，即可达到想要的目的。

C++代码实现如下：

class Solution {
public:
    void solve(vector<vector<char>> &board) {
        int height = board.size();
        if (height == 0) return;
        int width = board[0].size();
        if (width == 0) return;

        for (int i = 0; i < height; ++i) {
            if (board[i][0] == ‘O‘) BFSBoundary(board, i, 0);
            if (board[i][width - 1] == ‘O‘) BFSBoundary(board, i, width - 1);
        }

        for (int j = 1; j < width - 1; ++j) {
            if (board[0][j] == ‘O‘) BFSBoundary(board, 0, j);
            if (board[height - 1][j] == ‘O‘) BFSBoundary(board, height - 1, j);
        }

        for (int i = 0; i < height; ++i) {
            for (int j = 0; j < width; ++j) {
                if (board[i][j] == ‘O‘) board[i][j] = ‘X‘;
                if (board[i][j] == ‘#‘) board[i][j] = ‘O‘;
            }
        }
    }

private:
    void BFSBoundary(vector<vector<char>> &board, int h, int w) {
        int height = board.size();
        int width = board[0].size();

        queue<pair<int, int>> q;
        q.push(make_pair(h, w));
        board[h][w] = ‘#‘;

        while (!q.empty()) {
            pair<int, int> point = q.front();
            q.pop();

            pair<int, int> neighbors[4] = { { point.first + 1, point.second },
                                            { point.first - 1, point.second },
                                            { point.first, point.second + 1 },
                                            { point.first, point.second - 1 } };

            for (auto neig : neighbors) {
                if ((neig.first >= 0 && neig.first < height)
                    && (neig.second >= 0 && neig.second < width)
                    && (board[neig.first][neig.second] == ‘O‘)){
                    q.push(make_pair(neig.first, neig.second));
                    board[neig.first][neig.second] = ‘#‘;
                }
            }
        }
    }
};