[LeetCode]题解（python）：130-Surrounded Regions

题目来源：

　　https://leetcode.com/problems/surrounded-regions/

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题意分析：

　　给定给一个二维的板，这个板只包括‘X’和‘O’。将被‘X’包围的‘O’变成‘X’。比如：

X X X X
X O O X
X X O X
X O X X  
得到：

X X X X
X X X X
X X X X
X O X X

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题目思路：

　　从板的周围出发，从周围的‘O’出发深度搜索，能搜到的‘O’用visit记录他有没有访问过。最后将所有没有visit过的'O'变成‘X’.

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代码（python）：

 class Solution(object):
     def solve(self, board):
         """
         :type board: List[List[str]]
         :rtype: void Do not return anything, modify board in-place instead.
         """
         m = len(board)
         if m == 0:
             return
         n = len(board[0])
         visit = [[False for i in range(n)] for j in range(m)]
         def dfs(i,j):
             q = []
             q.append([i,j])
             while len(q) != 0:
                 tmp = q[0]
                 #print(tmp,visit[3][1],board[3][1])
                 q.pop(0)
                 #down,up,left,right
                 if tmp[0] - 1 > 0 and board[tmp[0] - 1][tmp[1]] == 'O' and visit[tmp[0]-1][tmp[1]] == False:
                     visit[tmp[0] -1][tmp[1]] = True
                     q.append([tmp[0] - 1,tmp[1]])
                 if tmp[0] + 1 < m and board[tmp[0] + 1][tmp[1]] == 'O' and visit[tmp[0]+1][tmp[1]] == False:
                     visit[tmp[0] +1][tmp[1]] = True
                     q.append([tmp[0] + 1,tmp[1]])
                 if tmp[1] - 1 > 0 and board[tmp[0]][tmp[1] - 1] == 'O' and visit[tmp[0]][tmp[1]-1] == False:
                     visit[tmp[0]][tmp[1] - 1] = True
                     q.append([tmp[0],tmp[1] - 1])
                 if tmp[1] + 1 < n and board[tmp[0]][tmp[1] + 1] == 'O' and visit[tmp[0]][tmp[1]+1] == False:
                     visit[tmp[0]][tmp[1]+1] = True
                     q.append([tmp[0],tmp[1]+1])
         for i in range(n):
             if visit[0][i] == False and board[0][i] == 'O':
                 visit[0][i] = True
                 dfs(0,i)
             if visit[m - 1][i] == False and board[m-1][i] == 'O':
                 visit[m-1][i] = True
                 dfs(m - 1,i)
         for j in range(m):
             if visit[j][0] == False and board[j][0] == 'O':
                 visit[j][0] = True
                 dfs(j,0)
             if visit[j][n - 1] == False and board[j][n - 1] == 'O':
                 visit[j][n - 1] = True
                 dfs(j,n - 1)
         for i in range(m):
             for j in range(n):
                 if board[i][j] == 'O' and not visit[i][j]:
                     board[i][j] = 'X'