leetcode1020

 1 class Solution(object):
 2     def __init__(self):
 3         self.cons = 0
 4         self.S = list()
 5 
 6     def dfs(self,m,n,v,A):
 7         while len(self.S) > 0:
 8             cur = self.S.pop(-1)
 9             cur_x = cur[0]
10             cur_y = cur[1]
11             v[cur_x][cur_y] =  1
12             self.cons += 1
13 
14             if cur_x > 0 and A[cur_x-1][cur_y]==1 and v[cur_x-1][cur_y] == 0:
15                 self.S.append([cur_x-1,cur_y])
16                 self.dfs(m,n,v,A)
17 
18             if cur_x < m - 1 and A[cur_x + 1][cur_y]==1 and v[cur_x + 1][cur_y]== 0:
19                 self.S.append([cur_x+1,cur_y])
20                 self.dfs(m,n,v,A)
21 
22             if cur_y > 0 and A[cur_x][cur_y-1]==1 and v[cur_x][cur_y-1]==0:
23                 self.S.append([cur_x,cur_y-1])
24                 self.dfs(m,n,v,A)
25 
26             if cur_y < n - 1 and A[cur_x][cur_y+1]==1 and v[cur_x][cur_y+1]==0:
27                 self.S.append([cur_x,cur_y+1])
28                 self.dfs(m,n,v,A)
29 
30 
31         return 0
32 
33     def numEnclaves(self, A: 'List[List[int]]') -> int:
34         row = len(A)
35         coloum = len(A[0])
36         visited =  [[0 for c in range(coloum)] for r in range(row)]
37         sumsA = 0
38         for i in range(row):
39             for j in range(coloum):
40                 sumsA += A[i][j]
41         if sumsA == 0:
42             return 0
43         
44         #first row and last row
45         for i in [0,row-1]:
46             for j in range(coloum):
47                 if A[i][j] == 0 or visited[i][j] == 1:
48                     continue
49                 else:
50                     self.S.append([i,j])
51                     self.dfs(row,coloum,visited,A)
52 
53         #first coloum and last coloum
54         for j in [0,coloum - 1]:
55             for i in range(row):
56                 if A[i][j] == 0 or visited[i][j] == 1:
57                     continue
58                 else:
59                     self.S.append([i,j])
60                     self.dfs(row,coloum,visited,A)
61 
62         return sumsA - self.cons

解决思路是使用DFS，为了提高效率，只扫描外边框（第一行，最后一行，第一列，最后一列），并使用一个数组记录是否已经访问。

posted on 2019-04-01 09:03 Sempron2800+ 阅读(115) 评论(0) 编辑收藏举报