[Leetcode] 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

先将所有外围的‘O’标记为‘.’，再将内部的‘O’标记成‘X’并且恢复外围的‘.’为‘O’。注意DFS时最外面的一圈已经在最开始处理过了，所以DFS时无需再进入了，否则会报内存错误。

DFS:

class Solution {
public:
    void dfs(vector<vector<char>> &board, int x, int y) {
        board[x][y] = '#';
        const int dx[4] = {0, 1, 0, -1};
        const int dy[4] = {1, 0, -1, 0};
        for (int i = 0; i < 4; ++i) {
            int xx = x + dx[i], yy = y + dy[i];
            if (xx > 0 && xx < board.size() - 1 && yy > 0 && yy < board[0].size() - 1 && board[xx][yy] == 'O') 
                dfs(board, xx, yy);
        }
    }
    void solve(vector<vector<char>>& board) {
        if (board.empty() || board[0].empty()) return;
        int N = board.size(), M = board[0].size();
        for (int i = 0; i < N; ++i) {
            if (board[i][0] == 'O') dfs(board, i, 0);
            if (board[i][M-1] == 'O') dfs(board, i, M-1);
        }
        for (int j = 0; j < M; ++j) {
            if (board[0][j] == 'O') dfs(board, 0, j);
            if (board[N-1][j] == 'O') dfs(board, N-1, j);
        }
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < M; ++j) {
                if (board[i][j] == '#') board[i][j] = 'O';
                else if (board[i][j] == 'O') board[i][j] = 'X';
            }
        }
        return;
    }
};

 

BFS:

class Solution {
public:
    void solve(vector<vector<char>>& board) {
        if (board.empty() || board[0].empty()) return;
        int N = board.size(), M = board[0].size();
        queue<pair<int, int>> que;
        for (int i = 0; i < N; ++i) {
            if (board[i][0] == 'O') que.push({i, 0});
            if (board[i][M-1] == 'O') que.push({i, M-1});
        }
        for (int j = 0; j < M; ++j) {
            if (board[0][j] == 'O') que.push({0, j});
            if (board[N-1][j] == 'O') que.push({N-1, j});
        }
        const int dx[4] = {0, 1, 0, -1};
        const int dy[4] = {1, 0, -1, 0};
        for (; !que.empty(); que.pop()) {
            auto u = que.front();
            int x = u.first, y = u.second;
            board[x][y] = '#';
            for (int i = 0; i < 4; ++i) {
                int xx = x + dx[i], yy = y + dy[i];
                if (xx >= 0 && xx < N && yy >= 0 && yy < M && board[xx][yy] == 'O') que.push({xx, yy});
            }
        }
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < M; ++j) {
                if (board[i][j] == '#') board[i][j] = 'O';
                else if (board[i][j] == 'O') board[i][j] = 'X';
            }
        }
        return;
    }
};