Lc200 小岛数量
每次碰到‘1’就用DFS遍历联通的1，把遍历的1标记为已访问; dfs单独写成一个函数
class Solution:
def _init_2d(self,row,col):
d= []
for i in range(row):
d.append([])
for j in range(col):
d[i].append(False)
return d
def dfs(self,grid,visit,i,j):
m,n = len(grid),len(grid[0])
if i<0 or i>=m or j<0 or j>=n or grid[i][j] == '0' or visit[i][j]:
return
visit[i][j] = True
self.dfs(grid,visit,i-1,j)
self.dfs(grid,visit,i+1,j)
self.dfs(grid,visit,i,j+1)
self.dfs(grid,visit,i,j-1)
def numIslands(self, grid: List[List[str]]) -> int:
if not grid or len(grid) == 0 or not grid[0] or len(grid[0])==0:
return 0
m,n = len(grid),len(grid[0])
visit = self._init_2d(m,n)
num = 0
for i in range(m):
for j in range(n):
if grid[i][j] == '0' or visit[i][j]:
continue
num+=1
self.dfs(grid,visit,i,j)
return num
