Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[0,0,1,0,0,0,0,1,0,0,0,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,1,1,0,1,0,0,0,0,0,0,0,0],
 [0,1,0,0,1,1,0,0,1,0,1,0,0],
 [0,1,0,0,1,1,0,0,1,1,1,0,0],
 [0,0,0,0,0,0,0,0,0,0,1,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,0,0,0,0,0,0,1,1,0,0,0,0]]

Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.
Note: The length of each dimension in the given grid does not exceed 50.

這個(gè)可以用DFS解決，不過，用C++或奸Java是能很快的解決的，但是，在leetcode上用python3我卻吃了不少虧，emmmmm

python3代碼：

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        seen = set() # 這個(gè)別放在class的外面了。。否則在第一個(gè)樣例能過，但在第二個(gè)樣例就會(huì)不過
        m = len(grid)
        n = len(grid[0])
        def dfs(r,c):
            if r < 0 or r >= m or c < 0 or c >= n or grid[r][c] == 0 or (r,c) in seen:
                return 0
            seen.add((r, c))
            return 1 + dfs(r+1, c) + dfs(r-1, c) + dfs(r, c+1) + dfs(r, c-1)
        ans = 0
        for r in range(m):
            for c in range(n):
                if grid[r][c] == 1:
                    ans = max(ans, dfs(r,c))
        return ans

起初我是想把dfs放在maxAreaOfIsland的外面，但是，我總是不能寫出正確的代碼，不過，當(dāng)把dfs放在里面的話，可以寫對(duì)了，不必加上self（比如self.dfs(...）)。