python--leetcode695. Max Area of Island

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.

题目意思是输入为1则是岛屿陆地，让你求最大的岛屿面积。

思路也就是递归去解：

class Solution(object):
    def maxAreaOfIsland(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        self.count=1
        self.max=0
        self.aa=[[0 for i in range(len(grid[0]))] for j in range(len(grid))]

        def search(grid,i,j):
            self.aa[i][j]=1
            if i>=1 :
                if grid[i-1][j]==1 and self.aa[i-1][j]==0:
                    self.count = self.count + 1
                    search(grid,i-1,j)

            if j>=1 :
                if grid[i][j-1]==1 and self.aa[i][j-1]==0:
                    self.count = self.count + 1
                    search(grid,i,j-1)
            if i<=len(grid)-2 :
                if grid[i+1][j]==1 and self.aa[i+1][j]==0:
                    self.count = self.count + 1
                    search(grid,i+1,j)
            if j<=len(grid[0])-2 :
                if grid[i][j+1]==1 and self.aa[i][j+1]==0:
                    self.count = self.count + 1
                    search(grid,i,j+1)
            if self.count>self.max:
                self.max=self.count

        for i in range(len(grid)):
            for j in range(len(grid[i])):
                self.count=1
                if grid[i][j]==1:
                    search(grid,i,j)

        return self.max

s=Solution()
print(s.maxAreaOfIsland([[1,1,1]]))

写的有点乱，aa是判断是否计算过该点的list。

看到一个更容易理解的解法：

   def maxAreaOfIsland(self, grid):
        m, n = len(grid), len(grid[0])

        def dfs(i, j):
            if 0 <= i < m and 0 <= j < n and grid[i][j]:
                grid[i][j] = 0
                return 1 + dfs(i - 1, j) + dfs(i, j + 1) + dfs(i + 1, j) + dfs(i, j - 1)
            return 0

        areas = [dfs(i, j) for i in range(m) for j in range(n) if grid[i][j]]
        return max(areas) if areas else 0