LEETCODE —— Surrounded Regions

Total Accepted: 43584 Total Submissions: 284350 Difficulty: Medium

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

反向思考，可以从四边开始找O，这些外面的O都找出来以后剩下的就是被包围的O了，非递归的BFS和DFS都可以AC(递归的很可能爆栈)，找出外面的O并替换为A，剩下的所有的O就是要找的，flip为X即可，再把A替换为O
AC之后的结果来看，DFS所需时间是BFS两倍。

BFS实现：

 class Solution(object):
     def solve(self, board):
         """
         :type board: List[List[str]]
         :rtype: void Do not return anything, modify board in-place instead.
         """
         if board==[]:
             return
         width = len(board[0])
         height = len(board)
         edgeNods=[]
         for column in range(width): #todo: all edgeNode
             edgeNode = (0, column)
             if board[0][column]=='O':
                 edgeNods.append(edgeNode)
         for line in range(height): #todo: all edgeNode
             edgeNode = (line, 0)
             if board[line][0]=='O':
                 edgeNods.append(edgeNode)
         for column in range(width): #todo: all edgeNode
             edgeNode = (height-1, column)
             if board[height-1][column]=='O':
                 edgeNods.append(edgeNode)
         for line in range(height): #todo: all edgeNode
             edgeNode = (line, width-1)
             if board[line][width-1]=='O':
                 edgeNods.append(edgeNode)

         stack = edgeNods
         while stack:
             nod = stack.pop(0)
             if board[nod[0]][nod[1]] == 'X' or board[nod[0]][nod[1]] == 'A' :
                 continue
             else:
                 board[nod[0]][nod[1]] = 'A'

             neighbours=[]
             neighbours.append((nod[0]-1, nod[1]))
             neighbours.append((nod[0]+1, nod[1]))
             neighbours.append((nod[0], nod[1]-1))
             neighbours.append((nod[0], nod[1]+1))
             for neiber in neighbours:
                 if neiber[0]<0 or neiber[0] >= height \
                     or neiber[1] <0  or neiber[1] >= width:
                     continue
                 if board[neiber[0]][neiber[1]] == 'X' or board[neiber[0]][neiber[1]] == 'A' :
                     continue
                 if (neiber in stack):
                     continue
                 stack.append(neiber)

         for column in range(width):
             for line in range(height):
                 if board[line][column] == 'O':
                     board[line][column]= "X"
         for column in range(width):
             for line in range(height):
                 if board[line][column] == 'A':
                     board[line][column]='O'