发表于 2024.04.15

• DFS往四个方面扩展面积即可，使用一个visited数组来维护已经访问的信息，如果遍历到已经visit的节点则不再统计。

  class Solution:
      moves = [
          (-1, 0),
          (0, -1),
          (1, 0),
          (0, 1)
      ]
  
      def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
          m = len(grid)
          n = len(grid[0])
          visited = [[False] * n for _ in range(m)]
  
          def dfs(x: int, y: int) -> int:
              nonlocal visited
  
              if grid[y][x] == 0:
                  return 0
              result = 1
              for move in self.moves:
                  nx = x + move[0]
                  ny = y + move[1]
                  if 0 <= nx < n and 0 <= ny < m and not visited[ny][nx]:
                      visited[ny][nx] = True
                      result += dfs(nx, ny)
              return result
  
          r = 0
          for i in range(m):
              for j in range(n):
                  if grid[i][j] and not visited[i][j]:
                      visited[i][j] = True
                      r = max(r, dfs(j, i))
          return r

• LC 题目链接