LeetCode130. Surrounded Regions

题目：

https://leetcode.com/problems/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

思路：

题意给一个OX组成的二维数组，把其中被X包围的O换成X。
嗯，先用dfs写个，不过这个题应该是喊用bfs吧，dfs过不了，有栈溢出。
如果O在二维数组的四边，肯定就不能算背包围。我们从四边往里搜索，如果四边上的位置有O，由这个位置的O往下找到的，与之相连的O，组成一块O区域，都不能被X包围。一边搜索，一边令这些O为#。在最后把二维数组中是#全部变成O，如果二维数组中还存在O，那么这些O都是被X包围的，应该换成X。

代码：

class Solution 
{
    public:
    int m,n;
    void dfs(vector<vector<char>> &board, int i, int j)
    {

        board[i][j]='#';
       if(i > 1 && board[i-1][j] == 'O')
        {
            dfs(board, i-1, j);
        }
        if(i < m-1 && board[i+1][j] == 'O')
        {
            dfs(board, i+1, j);
        }
       if(j > 1 && board[i][j-1] == 'O')
        {
            dfs(board, i, j-1);
        }
        if(j < n-1 && board[i][j+1] == 'O')
        {
            dfs(board, i, j+1);
        }
    }

    void solve(vector<vector<char> > &board) {
        m = board.size();
        if(m==0){
            return;
        }
        n = board[0].size();
        for(int j=0;j<n;j++)
        {
            if (board[0][j] == 'O')
            {
                dfs(board,0,j);
            }
            if (board[m-1][j] == 'O')
            {
                dfs(board,m-1,j);
            }
        }
        for(int i=1;i<m-1;i++)
        {
            if (board[i][0] == 'O')
            {
                dfs(board,i,0);
            }
            if (board[i][n-1] == 'O')
            {
                dfs(board,i,n-1);
            }
        }
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
            {
                if(board[i][j]=='O'){
                    board[i][j]='X';
                } 
                else if(board[i][j]=='#'){
                    board[i][j]='O';
                }
            } 
    }
};