[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）：

 1 class Solution(object):
 2     def solve(self, board):
 3         """
 4         :type board: List[List[str]]
 5         :rtype: void Do not return anything, modify board in-place instead.
 6         """
 7         m = len(board)
 8         if m == 0:
 9             return
10         n = len(board[0])
11         visit = [[False for i in range(n)] for j in range(m)]
12         def dfs(i,j):
13             q = []
14             q.append([i,j])
15             while len(q) != 0:
16                 tmp = q[0]
17                 #print(tmp,visit[3][1],board[3][1])
18                 q.pop(0)
19                 #down,up,left,right
20                 if tmp[0] - 1 > 0 and board[tmp[0] - 1][tmp[1]] == 'O' and visit[tmp[0]-1][tmp[1]] == False:
21                     visit[tmp[0] -1][tmp[1]] = True
22                     q.append([tmp[0] - 1,tmp[1]])
23                 if tmp[0] + 1 < m and board[tmp[0] + 1][tmp[1]] == 'O' and visit[tmp[0]+1][tmp[1]] == False:
24                     visit[tmp[0] +1][tmp[1]] = True
25                     q.append([tmp[0] + 1,tmp[1]])
26                 if tmp[1] - 1 > 0 and board[tmp[0]][tmp[1] - 1] == 'O' and visit[tmp[0]][tmp[1]-1] == False:
27                     visit[tmp[0]][tmp[1] - 1] = True
28                     q.append([tmp[0],tmp[1] - 1])
29                 if tmp[1] + 1 < n and board[tmp[0]][tmp[1] + 1] == 'O' and visit[tmp[0]][tmp[1]+1] == False:
30                     visit[tmp[0]][tmp[1]+1] = True
31                     q.append([tmp[0],tmp[1]+1])
32         for i in range(n):
33             if visit[0][i] == False and board[0][i] == 'O':
34                 visit[0][i] = True
35                 dfs(0,i)
36             if visit[m - 1][i] == False and board[m-1][i] == 'O':
37                 visit[m-1][i] = True
38                 dfs(m - 1,i)
39         for j in range(m):
40             if visit[j][0] == False and board[j][0] == 'O':
41                 visit[j][0] = True
42                 dfs(j,0)
43             if visit[j][n - 1] == False and board[j][n - 1] == 'O':
44                 visit[j][n - 1] = True
45                 dfs(j,n - 1)
46         for i in range(m):
47             for j in range(n):
48                 if board[i][j] == 'O' and not visit[i][j]:
49                     board[i][j] = 'X'
50 
51         

View Code

 

转载于:https://www.cnblogs.com/chruny/p/5307135.html