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[pull] master from youngyangyang04:master #374

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pull merged 3 commits into AlgorithmAndLeetCode:master from youngyangyang04:master
Nov 7, 2023
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116 changes: 60 additions & 56 deletions problems/0827.最大人工岛.md
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Original file line number Diff line number Diff line change
Expand Up @@ -284,64 +284,68 @@ class Solution {

### Python

```Python
class Solution(object):
# 可能的移动方向
DIRECTIONS = [(1, 0), (-1, 0), (0, 1), (0, -1)]

def exploreIsland(self, row, col, grid, visited, island_id):
"""
从给定的单元格开始,使用深度优先搜索探索岛屿。
"""
if (row < 0 or row >= len(grid) or col < 0 or col >= len(grid[0]) or
visited[row][col] or grid[row][col] == 0):
return 0

visited[row][col] = True
grid[row][col] = island_id
island_size = 1
for dr, dc in self.DIRECTIONS:
island_size += self.exploreIsland(row + dr, col + dc, grid, visited, island_id)
return island_size

def largestIsland(self, grid):
"""
通过最多将一个0更改为1,找到可以形成的最大岛屿的大小。
"""
rows, cols = len(grid), len(grid[0])
island_sizes = {}
island_id = 2 # 从2开始标记岛屿(因为0代表水,1代表未被发现的陆地)
is_all_land = True
visited = [[False] * cols for _ in range(rows)]

# 标记每个岛屿并存储其大小
for r in range(rows):
for c in range(cols):
```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:
is_all_land = False
elif not visited[r][c] and grid[r][c] == 1:
island_size = self.exploreIsland(r, c, grid, visited, island_id)
island_sizes[island_id] = island_size
island_id += 1

# 如果整个网格是陆地,则返回其大小
if is_all_land:
return rows * cols

# 计算可以通过将一个0更改为1来形成的最大岛屿
max_island_size = 0
for r in range(rows):
for c in range(cols):
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:
adjacent_islands = set()
for dr, dc in self.DIRECTIONS:
nr, nc = r + dr, c + dc
if 0 <= nr < rows and 0 <= nc < cols and grid[nr][nc] > 1:
adjacent_islands.add(grid[nr][nc])
new_island_size = sum(island_sizes[island] for island in adjacent_islands) + 1
max_island_size = max(max_island_size, new_island_size)

return max_island_size
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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