[leetcode] 695. Max Area of Island @ python

最新推荐文章于 2025-05-27 01:30:00 发布

原创最新推荐文章于 2025-05-27 01:30:00 发布 · 492 阅读

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本文介绍了一种使用深度优先搜索(DFS)算法解决二维数组中寻找最大岛屿面积问题的方法。通过遍历数组，遇到土地(1)时启动DFS，递归地计算岛屿面积并将其标记为已访问，最终返回所有岛屿中最大的面积。

原题

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.

解法

DFS. 先遍历grid, 当遇到1时, 求当前小岛的面积, 定义dfs函数返回小岛的面积, base case是当i, j超出了索引范围或者grid[i][j] 为0时, 返回0. 然后将走过的小岛标记为0, 返回 1+相邻小岛的面积.
Time: O(m*n)
Space: O(1)

代码

class Solution:
    def maxAreaOfIsland(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        def dfs(i, j):
            if 0 <= i < row and 0<= j < col and grid[i][j] == 1:
                grid[i][j] = 0
                return 1 + dfs(i-1, j) + dfs(i+1, j) + dfs(i, j-1) + dfs(i, j+1)
            return 0 
                    
        row, col = len(grid), len(grid[0])
        ans = 0
        for i in range(row):
            for j in range(col):
                if grid[i][j] == 1:
                    ans = max(ans, dfs(i, j))
                    
        return ans