Nov 7, 2023 · Oct 31, 2023 · Nov 7, 2023 · Nov 7, 2023 · Oct 31, 2023
diff --git a/problems/0827.最大人工岛.md b/problems/0827.最大人工岛.md
Original file line number	Diff line number	Diff line change
Expand Up		@@ -282,6 +282,71 @@ class Solution {
		}
		```

	### Python

	```python

	class Solution:
	def largestIsland(self, grid: List[List[int]]) -> int:
	visited = set() #标记访问过的位置
	m, n = len(grid), len(grid[0])
	res = 0
	island_size = 0 #用于保存当前岛屿的尺寸
	directions = [[0, 1], [0, -1], [1, 0], [-1, 0]] #四个方向
	islands_size = defaultdict(int) #保存每个岛屿的尺寸

	def dfs(island_num, r, c):
	visited.add((r, c))
	grid[r][c] = island_num #访问过的位置标记为岛屿编号
	nonlocal island_size
	island_size += 1
	for i in range(4):
	nextR = r + directions[i][0]
	nextC = c + directions[i][1]
	if (nextR not in range(m) or #行坐标越界
	nextC not in range(n) or #列坐标越界
	(nextR, nextC) in visited): #坐标已访问
	continue
	if grid[nextR][nextC] == 1: #遇到有效坐标,进入下一个层搜索
	dfs(island_num, nextR, nextC)

	island_num = 2 #初始岛屿编号设为2, 因为grid里的数据有0和1, 所以从2开始编号
	all_land = True #标记是否整个地图都是陆地
	for r in range(m):
	for c in range(n):
	if grid[r][c] == 0:
	all_land = False #地图里不全是陆地
	if (r, c) not in visited and grid[r][c] == 1:
	island_size = 0 #遍历每个位置前重置岛屿尺寸为0
	dfs(island_num, r, c)
	islands_size[island_num] = island_size #保存当前岛屿尺寸
	island_num += 1 #下一个岛屿编号加一
	if all_land:
	return m * n #如果全是陆地, 返回地图面积

	count = 0 #某个位置0变成1后当前岛屿尺寸
	#因为后续计算岛屿面积要往四个方向遍历,但某2个或3个方向的位置可能同属于一个岛,
	#所以为避免重复累加,把已经访问过的岛屿编号加入到这个集合
	visited_island = set() #保存访问过的岛屿
	for r in range(m):
	for c in range(n):
	if grid[r][c] == 0:
	count = 1 #把由0转换为1的位置计算到面积里
	visited_island.clear() #遍历每个位置前清空集合
	for i in range(4):
	nearR = r + directions[i][0]
	nearC = c + directions[i][1]
	if nearR not in range(m) or nearC not in range(n): #周围位置越界
	continue
	if grid[nearR][nearC] in visited_island: #岛屿已访问
	continue
	count += islands_size[grid[nearR][nearC]] #累加连在一起的岛屿面积
	visited_island.add(grid[nearR][nearC]) #标记当前岛屿已访问
	res = max(res, count)
	return res

	```

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