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.

 

class Solution(object):
    def maxAreaOfIsland(self, grid):
        h, w = len(grid), len(grid[0])

        def dfs(x, y):
            if 0 <= x < h and 0 <= y < w and grid[x][y]:
                grid[x][y] = 0
                return 1 + dfs(x-1, y) + dfs(x+1, y) + dfs(x, y-1) + dfs(x, y+1)
            return 0

        areas = [dfs(i, j) for i in range(h) for j in range(w) if gird[i][j]]
        return max(areas) if areas else 0

四方向dfs遍历，注意每次进入递归函数且确定是1后要置0，这样对于每一个island来说，第一次找到属于它的1时其它的全部都找到了，把他们标记为0后避免了重复查找。用for(h,w)在图中搜索尚未访问的1即可找到最大值。