leetcode 695. Max Area of Island

最新推荐文章于 2024-07-29 23:53:47 发布

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本文介绍了一种使用深度优先搜索(DFS)算法解决寻找二维数组中最大岛屿面积问题的方法。通过遍历矩阵，对每个岛屿进行DFS递归，计算其面积，并更新最大面积。示例代码展示了如何在Python中实现这一算法。

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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.

这个可以用DFS解决，不过，用C++或奸Java是能很快的解决的，但是，在leetcode上用python3我却吃了不少亏，emmmmm

python3代码：

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        seen = set() # 这个别放在class的外面了。。否则在第一个样例能过，但在第二个样例就会不过
        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，我总是不能写出正确的代码，不过，当把dfs放在里面的话，可以写对了，不必加上self（比如self.dfs(...）)。