695. 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 是由一些相邻的 1 (代表土地) 构成的组合，这里的「相邻」要求两个 1 必须在水平或者竖直方向上相邻。你可以假设 grid 的四个边缘都被 0（代表水）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。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/max-area-of-island/
Link：https://leetcode.com/problems/max-area-of-island/
宽度优先搜索
O(MN)
从一个点出发，进行宽度优先搜索，访问过的用0来标记, 注意宽度优先搜索使用队列，记得去重复
from collections import deque
class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
res = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
grid[i][j] = 0
area = self.helper(grid, i, j)
res = max(res, area)
return res
def helper(self, grid: List[List[int]], i: int, j: int) -> int:
x_vector = [-1, 0, 1, 0]
y_vector = [0, 1, 0, -1]
area = 0
queue = deque([(i,j)])
while len(queue) > 0:
x, y = queue.popleft()
area += 1
for i in range(4):
neighbor_x = x + x_vector[i]
neighbor_y = y + y_vector[i]
if 0 <= neighbor_x < len(grid) and 0 <= neighbor_y < len(grid[0]) and grid[neighbor_x][neighbor_y] == 1:
grid[neighbor_x][neighbor_y] = 0
queue.append((neighbor_x, neighbor_y))
return area
深度优先搜索
O(MN)
从一个点出发，进行深度优先搜索，访问过的用0来标记, 注意宽度优先搜索使用栈，记得去重复
当然和可以用递归，递归优点通常是代码容易实现，方便回溯。缺点就是容易栈溢出。
⚠️注意，方向数组也可以用一维数组表示
class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
res = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
grid[i][j] = 0
area = self.helper(grid, i, j)
res = max(res, area)
return res
def helper(self, grid: List[List[int]], i: int, j: int) -> int:
vector = [-1, 0, 1, 0, -1]
area = 0
stack = [(i,j)]
while len(stack) > 0:
x, y = stack.pop()
area += 1
for k in range(4):
neighbor_x = x + vector[k]
neighbor_y = y + vector[k + 1]
if 0 <= neighbor_x < len(grid) and 0 <= neighbor_y < len(grid[0]) and grid[neighbor_x][neighbor_y] == 1:
grid[neighbor_x][neighbor_y] = 0
stack.append((neighbor_x, neighbor_y))
return area
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