695. Max Area of Island@python

Posted 2021-01-09 Chim

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.

 题意:在一个二维数组中,找出能够连接在一起的数字1的最大长度

思路:DFS

代码:

class Solution(object):
    def maxAreaOfIsland(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        res = 0
        m = len(grid)
        n = len(grid[0])
        for i in range(m):    # 遍历数组
            for j in range(n):
                if grid[i][j] == 1:
                    res = max(res, self.dfs(grid, i, j))
        return res
        
    
    def dfs(self, grid, i, j):
        m = len(grid)
        n = len(grid[0])
        if i < 0 or i > m-1 or j < 0 or j > n-1             or grid[i][j] == 0:
            return 0
        count = 1
        grid[i][j] = 0   # 将值变为0,防止重复计数或者递归栈溢出
        count += self.dfs(grid, i+1, j) +                 self.dfs(grid, i-1, j) +                 self.dfs(grid, i, j-1) +                 self.dfs(grid, i, j+1) 
        
        return count 

时间复杂度: O(mn)

空间复杂度:O(1)