添加0827.最大人工岛Python3版本

This commit is contained in:
Xiong Gu
2023-10-30 21:09:36 -04:00
parent d5b2f683d1
commit c35a886d35

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@ -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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