leetcode1020

 class Solution(object):
     def __init__(self):
         self.cons = 0
         self.S = list()

     def dfs(self,m,n,v,A):
         while len(self.S) > 0:
             cur = self.S.pop(-1)
             cur_x = cur[0]
             cur_y = cur[1]
             v[cur_x][cur_y] =  1
             self.cons += 1

             if cur_x > 0 and A[cur_x-1][cur_y]==1 and v[cur_x-1][cur_y] == 0:
                 self.S.append([cur_x-1,cur_y])
                 self.dfs(m,n,v,A)

             if cur_x < m - 1 and A[cur_x + 1][cur_y]==1 and v[cur_x + 1][cur_y]== 0:
                 self.S.append([cur_x+1,cur_y])
                 self.dfs(m,n,v,A)

             if cur_y > 0 and A[cur_x][cur_y-1]==1 and v[cur_x][cur_y-1]==0:
                 self.S.append([cur_x,cur_y-1])
                 self.dfs(m,n,v,A)

             if cur_y < n - 1 and A[cur_x][cur_y+1]==1 and v[cur_x][cur_y+1]==0:
                 self.S.append([cur_x,cur_y+1])
                 self.dfs(m,n,v,A)

         return 0

     def numEnclaves(self, A: 'List[List[int]]') -> int:
         row = len(A)
         coloum = len(A[0])
         visited =  [[0 for c in range(coloum)] for r in range(row)]
         sumsA = 0
         for i in range(row):
             for j in range(coloum):
                 sumsA += A[i][j]
         if sumsA == 0:
             return 0

         #first row and last row
         for i in [0,row-1]:
             for j in range(coloum):
                 if A[i][j] == 0 or visited[i][j] == 1:
                     continue
                 else:
                     self.S.append([i,j])
                     self.dfs(row,coloum,visited,A)

         #first coloum and last coloum
         for j in [0,coloum - 1]:
             for i in range(row):
                 if A[i][j] == 0 or visited[i][j] == 1:
                     continue
                 else:
                     self.S.append([i,j])
                     self.dfs(row,coloum,visited,A)

         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。