class Solution:
def pondSizes(self, land: List[List[int]]) -> List[int]:
if len(land)==0:return []
directions=[(-1,0),(1,0),(0,-1),(0,1),(-1,1),(-1,-1),(1,-1),(1,1)]
res,m,n=[],len(land),len(land[0])
for i in range(m):
for j in range(n):
if land[i][j]==0:
queue=[(i,j)]
land[i][j]=1
cnt=1
while queue:
next_queue=[]
for old_x,old_y in queue:
for direction in directions:
new_x,new_y=old_x+direction[0],old_y+direction[1]
if 0<=new_x<m and 0<=new_y<n and land[new_x][new_y]==0:
land[new_x][new_y]=1
cnt+=1
next_queue.append((new_x,new_y))
queue=next_queue
res.append(cnt)
res.sort()
return res BFS计算有多少个8方向连接的岛屿,并且按照岛屿的面积从小到大输出一个列表。

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