时间复杂度： O(n)

空间复杂度：O(n)

题意解析：

就是把所有被“x”包围的“o”替换成 x。所谓 “包围” 是指 上下左右 四个方向，不包括斜上，斜下。。。

算法思路：

没有被“x” 围住： 就是 那一组联通的“o“ 连接到边界了，只要把连接到边界的 ”o“ 替换成*，其他的o就替换成x，最后再把*还原成O

在把连接到边界的O替换成* 用的是bfs，具体代码如下，time：52ms

/*
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
*/

class Solution {
public:
	void solve(vector<vector<char>> &board) {
		int height = board.size(), width;
		if (height == 0)
			return;
		width = board[0].size();
		//search the 2 colum (the first colum and  the last colum)
		for (int i = 0; i < height; i++){
			if (board[i][0] == 'O')
				bfs(i, 0, board);
			if (board[i][width - 1] == 'O')
				bfs(i, width - 1, board);
		}
		//search the first and last rows
		for (int i = 1; i < width - 1; i++){
			if (board[0][i] == 'O')
				bfs(0, i, board);
			if (board[height - 1][i] == 'O')
				bfs(height - 1, i, board);
		}
		for (int i = 0; i < height; i++){
			for (int j = 0; j < width; j++){
				if (board[i][j] == '*'){
					board[i][j] = 'O';
				}
				if (board[i][j] == 'O'){
					board[i][j] = 'X';
				}
			}
		}
	}
	

	private:
	struct position{ 
		int x;
		int y;
	};
	void bfs(int x, int y, vector<vector<char>> &board){
		queue<position>	q_pos;
		visit(x, y, board, q_pos);
		while (!q_pos.empty()){
			auto next_pos = q_pos.front();
			q_pos.pop();
			visit(x - 1, y, board, q_pos);
			visit(x + 1, y, board, q_pos);
			visit(x, y + 1, board, q_pos);
			visit(x, y - 1, board, q_pos);
		}
	}
	void visit(int x, int y, vector<vector<char>> &board, queue<position> &q_pos){
		if (x < 0 || x >= board.size() || y < 0 || y >= board[0].size() || board[x][y] == 'X')
			return;
		if (board[x][y] == 'O'){
			board[x][y] = '*';
			position p;
			p.x = x;
			p.y = y;
			q_pos.push(p);
		}
	}
};

leetcode——surronded regions