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