Surrounded Regions

Problem:

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

尝试了几种扫描整个平面的方法，都失败了。主要原因是二维扫描的顺序不能向四个方向发散，总会有遗漏的死角。尝试一种递归的方法，出现了StackOverFlow 的runtime error。看有评论说用栈模拟递归能解决这个问题。最后用了下面这个队列进行BFS的方法，终于AC了。

Solution:

public class Solution {
public void solve(char[][] board) {
        if(board==null||board.length==0||board[0].length==0)
         return ;
       
        LinkedList<BoardPoint> pQueue = new LinkedList<>();
       int len = board.length;
       int wid = board[0].length;
        
        for(int i=0;i<len;i++)
        {
            if(board[i][wid-1]=='O')
            {
                board[i][wid-1] = 'N';
                if(wid>1&&board[i][wid-2]=='O')
                {
                    board[i][wid-2] = 'N';
                    pQueue.add(new BoardPoint(i, wid-2));
                }
            }
            if(board[i][0]=='O')
            {
                board[i][0] = 'N';
                if(wid>1&&board[i][1]=='O')
                {
                    board[i][1] = 'N';
                    pQueue.add(new BoardPoint(i, 1));
                }
            }
         }
        
        for(int i=0;i<wid;i++)
        {
             if(board[0][i]=='O')
             {
                board[0][i] = 'N';
                if(len>1&&board[1][i]=='O')
                {
                    board[1][i] = 'N';
                    pQueue.add(new BoardPoint(1, i));
                }
             }
             if(board[len-1][i]=='O')
            {
                board[len-1][i] = 'N';
                if(len>1&&board[len-2][i]=='O')
                {
                    board[len-2][i] = 'N';
                    pQueue.add(new BoardPoint(len-2, i));
                }
            }
        }
        
        while(!pQueue.isEmpty())
        {
             BoardPoint p = pQueue.poll();
             if(p.x<len-2&&board[p.x+1][p.y]=='O')
             {
                  board[p.x+1][p.y] = 'N';
                  pQueue.addLast(new BoardPoint(p.x+1, p.y));
             }
             if(p.x>1&&board[p.x-1][p.y]=='O')
             {
                 board[p.x-1][p.y] = 'N';
                 pQueue.addLast(new BoardPoint(p.x-1, p.y));
             }
             if(p.y<wid-2&&board[p.x][p.y+1]=='O')
             {
                 board[p.x][p.y+1] = 'N';
                 pQueue.addLast(new BoardPoint(p.x, p.y+1));
             }
             if(p.y>1&&board[p.x][p.y-1]=='O')
             {
                 board[p.x][p.y-1] = 'N';
                 pQueue.addLast(new BoardPoint(p.x, p.y-1));
             }
        }
        
        for(int i=0;i<len;i++)
        {
             for(int j=0;j<wid;j++)
             {
                 if(board[i][j]=='N')
                     board[i][j] = 'O';
                 else if(board[i][j]=='O')
                     board[i][j] = 'X';
             }
        }
    }
}


class BoardPoint
{
int x,y;
public BoardPoint(int x,int y)
{
this.x = x;
this.y = y;
}
}