给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
岛屿总是被水包围，并且每座岛屿只能由水平方向或竖直方向上相邻的陆地连接形成。
此外，你可以假设该网格的四条边均被水包围。
示例 1:
输入:
[
['1','1','1','1','0'],
['1','1','0','1','0'],
['1','1','0','0','0'],
['0','0','0','0','0']
]
输出: 1
使用DFS遍历每个值==1的岛屿位置，一次从头的DFS把值==1的岛屿转化为‘0’下次不再重复遍历，这样DFS的次数就是岛屿的数量.
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid:
return 0
n=len(grid)
m=len(grid[0])
def dfs(grid,i,j):
if 0<=i<=n-1 and 0<=j<=m-1 and grid[i][j]=='1':
grid[i][j]='0'
dfs(grid,i+1,j)
dfs(grid,i,j-1)
dfs(grid,i-1,j)
dfs(grid,i,j+1)
count=0
for i in range(n):
for j in range(m):
if grid[i][j]=='1':
dfs(grid,i,j)
count+=1
return count
使用BFS是类似的相同的思路，BFS的次数是岛屿数量
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
n=len(grid)
if n==0:
return 0
m=len(grid[0])
count=0
for i in range(n):
for j in range(m):
if grid[i][j]=='1':
count+=1
grid[i][j]='0'
q=collections.deque([(i,j)])
while q:
row,col=q.popleft()
for x,y in [(row-1,col),(row,col+1),(row+1,col),(row,col-1)]:
if 0<=x使用并查集的思路更加清晰，可能实现并查集的三个操作略微麻烦一些
class UnionFind:
def __init__(self,grid):
n=len(grid)
m=len(grid[0])
self.parent=[-1]*(n*m)
self.value=[1]*(n*m)
self.count=0
for i in range(n):
for j in range(m):
if grid[i][j]=='1':
self.parent[i*m+j]=i*m+j
self.count+=1
def find(self,i):
if self.parent[i]!=i:
self.parent[i] =self.find(self.parent[i])
return self.parent[i]
def union(self,x,y):
rx=self.find(x)
ry=self.find(y)
if rx!=ry:
if self.value[rx]self.value[ry]:
self.parent[rx]=ry
self.value[ry]+=self.value[rx]
if self.value[rx]==self.value[ry]:
self.value[rx]+=1
self.parent[rx]=ry
self.count-=1
def getCount(self):
return self.count
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
n=len(grid)
if n==0:
return 0
m=len(grid[0])
uf=UnionFind(grid)
for i in range(n):
for j in range(m):
if grid[i][j]=='1':
grid[i][j]='0'
for x,y in [(i-1,j),(i,j+1),(i+1,j),(i,j-1)]:
if 0<=x