130. Surrounded Regions

这道题只要把与边缘相连的‘O’全部换成‘V’，然后将所有的除‘V’之外的字母全部换成‘X’,最后再讲所有的‘V’换成‘O’。 方法的就是将DFS与边缘相连的‘O’, 或者BFS与边缘相连的‘O’。但是递归版本的DFS栈溢出，所有用栈的DFS即可。
1. 超时版本

class Solution {
public:
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++)
        for(int j = 0; j < m; j++)
            if(i == 0 || i == n-1 || j == 0 || j == m-1)
                dfs(board, i, j);
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            board[i][j] = (board[i][j] == 'V') ? 'O' : 'X';
}

void dfs(vector<vector<char>> &board, int row, int col)
{
    int n = board.size(), m = board[0].size();
    if(row < 0 || row >= n || col < 0 || col >= m) return;
    if(board[row][col] != 'O') return;
    board[row][col] = 'V';
    dfs(board, row+1, col);
    dfs(board, row-1, col);
    dfs(board, row, col+1);
    dfs(board, row, col-1);

    }
};

1. 改进版本 用栈

 class Solution {
    public:
        typedef vector<vector<char> > BOARDTYPE;

        void solve(BOARDTYPE &board) {
            if (board.empty() || board[0].empty()) return;
            int N = board.size(), M = board[0].size();
            for (int i = 0; i < N; ++i)
                for (int j = 0; j < M; ++j)
                    if (i == 0 || j == 0 || i == N-1 || j == M-1)
                        dfs(board, i, j); // you may call dfs or bfs here!
            for (int i = 0; i < N; ++i)
                for (int j = 0; j < M; ++j)
                    board[i][j] = (board[i][j] == 'V') ? 'O' : 'X';
        }

        void dfs(vector<vector<char>> &board, int row, int col)
        {
            int n = board.size(), m = board[0].size();
            if(row < 0 || row >= n || col < 0 || col >= m) return;
            if(board[row][col] != 'O') return;
            stack<pair<int, int>> s;
            s.push(make_pair(row, col));
            while(!s.empty())
            {
                row = s.top().first;
                col = s.top().second;
                s.pop();
                board[row][col] = 'V';
                if(row+1 < n && board[row+1][col] == 'O')
                    s.push(make_pair(row+1, col));
                if(row-1 >= 0 && board[row-1][col] == 'O')
                    s.push(make_pair(row-1, col));
                if(col+1 < m && board[row][col+1] == 'O')
                    s.push(make_pair(row, col+1));
                if(col-1 >= 0 && board[row][col-1] == 'O')
                    s.push(make_pair(row, col-1));

            }
        }

    };  

1. BFS方法

 class Solution {
    public:
        typedef vector<vector<char> > BOARDTYPE;

        void solve(BOARDTYPE &board) {
            if (board.empty() || board[0].empty()) return;
            int N = board.size(), M = board[0].size();
            queue<pair<int,int>> q;

            for (int i = 0; i < N; ++i){
                if(board[i][M-1]=='O' ){
                    q.push(make_pair(i,M-1));
                    board[i][M-1]='V';     
                }    
                if(board[i][0]=='O'){
                    q.push(make_pair(i,0));
                    board[i][0]='V';
                }
            }
            for (int j = 0; j < M; ++j){
                if(board[0][j]=='O'){
                    q.push(make_pair(0,j));    
                    board[0][j]='V';
                }     
                if( board[N-1][j]=='O'){
                    q.push(make_pair(N-1,j));
                    board[N-1][j]='V';
                }

            }
            bfs(board,q);
            for (int i = 0; i < N; ++i)
                for (int j = 0; j < M; ++j)
                    board[i][j] = (board[i][j] == 'V') ? 'O' : 'X';
        }

         void bfs(BOARDTYPE &board, queue<pair<int,int>> &q) {
            int N = board.size(), M = board[0].size();

            while(!q.empty()){
                int row = q.front().first;
                int col = q.front().second;
                q.pop();


                if(row+1<N && board[row+1][col]=='O'){
                    q.push(make_pair(row+1,col));
                    board[row+1][col]='V';
                }
                if(row-1>=0 && board[row-1][col]=='O'){
                    q.push(make_pair(row-1,col));
                    board[row-1][col]='V';
                }
                if(col+1<M && board[row][col+1]=='O'){
                    q.push(make_pair(row,col+1));
                    board[row][col+1]='V';
                }
                if(col-1<N && board[row][col-1]=='O'){
                    q.push(make_pair(row,col-1));
                    board[row][col-1]='V';
                }
            }//end while
        }
    };