LeetCode题解：Surrounded Regions

Surrounded Regions

Total Accepted: 1947 Total Submissions: 13017

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

思路：

搜索棋盘上的'O'，找到一个之后做一个DFS或者BFS均可，做的时候随时检查是否到达棋盘的边缘。到的话就不转化成‘X'，否则按照搜索记录把棋盘上的网格转换成’X'。

题解：

class Solution {
public:

    void navigate_and_fill (vector<vector<char>>& board,
                            int R, int C, int M, int N)
    {
        vector<pair<int, int>> nav_history;
        
        bool surrounded = true;
        queue<pair<int, int>> nav_queue;
        nav_queue.push (make_pair (R, C));
        board[R][C] = 'o';
        while (!nav_queue.empty())
        {
            auto q = nav_queue.front();
            nav_history.push_back(q);
            nav_queue.pop();
            
            if (q.first == 0 || q.first == M - 1 ||
                    q.second == 0 || q.second == N - 1)
                surrounded = false;
    
            if (q.first != 0 && board[q.first - 1][q.second] == 'O')
            {
                board[q.first - 1][q.second] = 'o';
                nav_queue.push (make_pair (q.first - 1, q.second));
            }
                
            if (q.first != M - 1 && board[q.first + 1][q.second] == 'O')
            {
                board[q.first + 1][q.second] = 'o';
                nav_queue.push (make_pair (q.first + 1, q.second));
            }
                
            if (q.second != 0 && board[q.first][q.second - 1] == 'O')
            {
                board[q.first][q.second - 1] = 'o';
                nav_queue.push (make_pair (q.first, q.second - 1));
            }
            
            if (q.second != N - 1 && board[q.first][q.second + 1] == 'O')
            {
                board[q.first][q.second + 1] = 'o';
                nav_queue.push (make_pair (q.first , q.second + 1));
            }
        }
        
        if (surrounded)
            for (auto & n : nav_history)
                board[n.first][n.second] = 'X';
    }
    
    void solve (vector<vector<char>>& board)
    {
        const int M = board.size();
        if (M == 0)
            return;
        
        const int N = board[0].size();
        if (N == 0)
            return;
        
        for (int R = 0; R < M; ++R)
            for (int C = 0; C < N; ++C)
                if (board[R][C] == 'O')
                    navigate_and_fill (board, R, C, M, N);
    
        for (auto & r : board)
            for (auto & c : r)
                if (c == 'o') c = 'O';
    }

};