LeetCode -- Surrounded Regions

题目描述：
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


在由'X'和'O'组成的矩阵中，找出所有被'X'包围的'O'，将它们替换为'X'。


思路：
1.本题可以DFS也可以BFS，但是DFS速度会慢一些，无法通过测试数据。无论哪种方式，就是从四个边的'O'作为切入点n（如果BFS，需要先将四个边的'O'入队列），遍历时找到n的上下左右邻居，分别进行标记（假设标记为A），凡是标记为A的点即为：不能被围的'O'，以这些邻居为中心继续下一次标记。
2.然后对board做一个遍历，把剩余的'O'赋值为'X'，而'A'赋值为'O'即可。


实现代码：

public class Solution {    public void Solve(char[,] board)     {       var row = board.GetLength(0);    var col = board.GetLength(1);        if(row < 2 || col < 2){    return;    }        // try go from left & right boundary    var q = new Queue<Pos>();    for(var i = 0;i < row; i++){    if(board[i , 0] == 'O'){    q.Enqueue(new Pos(i , 0));    }    if(board[i , col - 1] == 'O'){    q.Enqueue(new Pos(i , col - 1));    }    }        // try go from top & down boundary    for(var i = 0;i < col; i++){    if(board[0,i] == 'O'){    q.Enqueue(new Pos(0 , i));    }    if(board[row - 1 , i] == 'O'){    q.Enqueue(new Pos(row - 1 , i));    }    }    Bfs(ref board, row, col , q);        // restore 'A' to 'O'    // mark 'O' to 'X'    for(var i = 0;i < row; i++){    for(var j = 0;j < col; j++){    if(board[i,j] == 'O'){    board[i,j] = 'X';    }    if(board[i,j] == 'A'){    board[i,j] = 'O';    }    }    }    }private void Bfs(ref char[,] board, int rowLen, int colLen, Queue<Pos> q){if(q.Count == 0){return;}var q1 = new Queue<Pos>();while(q.Count > 0){var p = q.Dequeue();board[p.row, p.col] = 'A';// move upif(p.row > 0 && board[p.row - 1, p.col] == 'O'){q1.Enqueue(new Pos(p.row - 1, p.col));}// move downif(p.row < rowLen - 1 && board[p.row + 1, p.col] == 'O'){q1.Enqueue(new Pos(p.row + 1, p.col));}// move rightif(p.col < colLen - 1 && board[p.row, p.col + 1] == 'O'){q1.Enqueue(new Pos(p.row, p.col + 1));}// move leftif(p.col > 0 && board[p.row, p.col - 1] == 'O'){q1.Enqueue(new Pos(p.row, p.col - 1));}}Bfs(ref board, rowLen, colLen, q1);}class Pos{public Pos(int r, int c){row = r;col = c;}public int row;public int col;}}