[LeetCode] Surrounded Regions

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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

用bfs没问题

class Solution {public:    int m,n;    queue<int> q;    void add(int x,int y,vector<vector<char>> &board){        if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O'){            board[x][y] = 'q';            q.push(x*n + y);        }    }    void bfs(int x,int y,vector<vector<char>> &board){        add(x,y,board);        while(!q.empty()){            int p = q.front();            q.pop();            int px = p / n,py = p % n;            add(px - 1,py,board);            add(px + 1,py,board);            add(px,py - 1,board);            add(px,py + 1,board);        }    }    void solve(vector<vector<char>> &board) {        if(board.empty()) return;        m = board.size();        n = board[0].size();        for(int i = 0;i < n;i ++){            bfs(0,i,board);            bfs(m - 1,i,board);        }        for(int i = 0;i < m;i ++){            bfs(i,0,board);            bfs(i,n - 1,board);        }        for(int i = 0;i < m;i ++){            for(int j = 0;j < n;j ++){                board[i][j] = (board[i][j] == 'q' ? 'O' : 'X');            }        }    }};

用dfs会超时

class Solution {public:    int m,n;    void dfs(int x,int y,vector<vector<char>> &board){        if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O'){            board[x][y] = 'q';            dfs(x - 1,y,board);            dfs(x + 1,y,board);            dfs(x,y - 1,board);            dfs(x,y + 1,board);        }    }    void solve(vector<vector<char>> &board) {        if(board.empty()) return;        m = board.size();        n = board[0].size();        for(int i = 0;i < n;i ++){            dfs(0,i,board);            dfs(m - 1,i,board);        }        for(int i = 0;i < m;i ++){            dfs(i,0,board);            dfs(i,n - 1,board);        }        for(int i = 0;i < m;i ++){            for(int j = 0;j < n;j ++){                board[i][j] = board[i][j] == 'q' ? 'O' : 'X';            }        }    }};

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