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.
解法1:dfs
class Solution(object): def maxAreaOfIsland(self, grid): """ :type grid: List[List[int]] :rtype: int """ # reset 1 to 0 use dfs, and return the ones numer row = len(grid) col = len(grid[0]) ans = 0 def dfs(grid, i, j): if i==row or j==col or i<0 or j<0 or grid[i][j]==0: return 0 res = 1 grid[i][j] = 0 res += dfs(grid, i+1, j) res += dfs(grid, i-1, j) res += dfs(grid, i, j-1) res += dfs(grid, i, j+1) return res for i in range(0, row): for j in range(0, col): if grid[i][j] == 1: ans = max(ans, dfs(grid, i, j)) return ans
解法2:BFS，必须在if里将1reset为0，否则有重复：
class Solution(object): def maxAreaOfIsland(self, grid): """ :type grid: List[List[int]] :rtype: int """ # reset 1 to 0 use bfs, and return the ones numer row = len(grid) col = len(grid[0]) ans = 0 def bfs(grid, i, j): assert grid[i][j] == 1 q = [(i,j)] grid[i][j]=0 res = 1 while q: q2 = [] for i,j in q: if i+1=0 and grid[i-1][j]==1: grid[i-1][j]=0 q2.append((i-1,j)) res += 1 if j+1=0 and grid[i][j-1]==1: grid[i][j-1]=0 q2.append((i,j-1)) res += 1 q = q2 return res for i in range(0, row): for j in range(0, col): if grid[i][j] == 1: ans = max(ans, bfs(grid, i, j)) return ans