[leetcode] 695. Max Area of Island

本文主要是介绍[leetcode] 695. Max Area of Island，对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

题目

You are given an m x n binary matrix grid. 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.

The area of an island is the number of cells with a value 1 in the island.

Return the maximum area of an island in grid. If there is no island, return 0.

Example 1:

Input: grid = [[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]]
Output: 6
Explanation: The answer is not 11, because the island must be connected 4-directionally.

Example 2:

Input: grid = [[0,0,0,0,0,0,0,0]]
Output: 0

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 50
• grid[i][j] is either 0 or 1.

思路

dfs，自上而下将问题分解为求当前位置与上下左右位置面积的子问题，然后相加这些求得的面积。

代码

python版本：

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        def dfs(i, j):
            if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == 0:
                return 0
            grid[i][j] = 0
            return dfs(i-1, j)+dfs(i+1, j)+dfs(i, j-1)+dfs(i, j+1)+1
        res = [dfs(i, j) for i in range(len(grid)) for j in range(len(grid[0]))]
        return max(res)

这篇关于[leetcode] 695. Max Area of Island的文章就介绍到这儿，希望我们推荐的文章对大家有所帮助，也希望大家多多支持为之网！