LeetCode695--岛屿的最大面积

 '''
 岛屿的最大面积
 给定一个包含了一些 0 和 1的非空二维数组 grid , 一个 岛屿 是由四个方向 (水平或垂直) 的 1 (代表土地) 构成的组合。你可以假设二维矩阵的四个边缘都被水包围着。
 找到给定的二维数组中最大的岛屿面积。(如果没有岛屿，则返回面积为0。)
 示例 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]]
 对于上面这个给定矩阵应返回 6。注意答案不应该是11，因为岛屿只能包含水平或垂直的四个方向的‘1’。

 示例 2:
 [[0,0,0,0,0,0,0,0]]
 对于上面这个给定的矩阵, 返回 0。
 注意: 给定的矩阵grid 的长度和宽度都不超过 50。
 '''

 class Solution:
     def maxAreaOfIsland(self, grid):
         """
         :type grid: List[List[int]]
         :rtype: int
         """
         mx = 0
         for i in range(len(grid)):
             for j in range(len(grid[i])):
                 if grid[i][j] == 1:
                     num = self.deepSearch(grid, i, j)
                     mx = max(num, mx)
         return mx

     def deepSearch(self, g, i, j):
         g[i][j] = 0
         x = len(g)
         y = len(g[0])
         c = 1
         if i - 1 >= 0 and g[i - 1][j]:  # 上
             c = c + self.deepSearch(g, i - 1, j)
         if j + 1 < y and g[i][j + 1]:  # 右
             c = c + self.deepSearch(g, i, j + 1)
         if i + 1 < x and g[i + 1][j]:  # 下
             c = c + self.deepSearch(g, i + 1, j)
         if j - 1 >= 0 and g[i][j - 1]:  # 左
             c = c + self.deepSearch(g, i, j - 1)
         return c

 if __name__ == '__main__':
     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]]
     ret = Solution().maxAreaOfIsland(grid)
     print(ret)