题解 | #岛屿数量#
岛屿数量
https://www.nowcoder.com/practice/0c9664d1554e466aa107d899418e814e
#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
# 判断岛屿数量
# @param grid char字符型二维数组
# @return int整型
#
dx = [1,-1,0,0]
dy = [0,0,1,-1]
class Solution:
def __init__(self) -> None:
self.flag = {}
self.grid = None
self.count = []
self.delone = {}
self.vis = {}
def ok(self,x,y):
n = self.n
m = self.m
if x<0 or x>n-1 or y<0 or y>m-1:
return False
elif (x,y) in self.vis.keys():
return False
else:
return True
def search(self,this,nowflag):
if this in self.vis.keys():
return 0
val = self.grid[this[0]][this[1]]
self.vis[this]=1
if val=='0':
return 0
if this in self.flag.keys():
thisflag = self.flag[this]
minc = min(thisflag,nowflag)
delc = max(thisflag,nowflag)
self.delone[delc]=minc
self.count.remove(delc)
thisflag = minc
self.flag[this]=thisflag
else:
self.flag[this] = nowflag
thisflag = self.flag[this]
if nowflag not in self.count:
self.count.append(nowflag)
for _ in range(4):
x = this[0] +dx[_]
y = this[1] +dy[_]
if self.ok(x,y):
back = self.search((x,y),thisflag)
return 1
def solve(self , grid: List[List[str]]) -> int:
# write code here
self.grid = grid
self.n = len(grid)
self.m = len(grid[0])
nowflag = 1
for i in range(self.n):
for j in range(self.m):
back = self.search((i,j),nowflag)
if back == 1:
nowflag+=1
return len(self.count)