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 XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX 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';    }};