[LeetCode] Challenge log 695

Posted on | Edited on |
695. Max Area of Island

Recursive solution. Remember to check the seen slot.

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:

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[[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 

1
6

Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

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

Given the above grid, return 

1
0

Note: The length of each dimension in the given grid does not exceed 50.

Soulution:
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class Solution:
    def maxAreaOfIsland(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        nrow = len(grid)
        ncol = len(grid[0])
        def countArea(x, y):
            grid[x][y] = 0 
            cur = 1
            if x + 1 < nrow and grid[x + 1][y] == 1:
                cur += countArea(x + 1, y)
            if x - 1 >= 0 and grid[x - 1][y] == 1:
                cur += countArea(x - 1, y)
            if y + 1 < ncol and grid[x][y + 1] == 1:
                cur += countArea(x, y + 1)
            if y - 1 >= 0 and grid[x][y - 1] == 1:
                cur += countArea(x, y - 1)
            return cur
        
        res = 0
        for i in range(nrow):
            for j in range(ncol):
                if grid[i][j] == 1:
                    res = max(res, countArea(i,j))
        return res
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