Python中的孤岛数量
假设我们有一个网格，有几个0和几个1。我们必须计算岛屿的数量。岛屿是一个被水包围的地方，是通过水平或垂直连接相邻的土地而形成的。我们可以假设网格的所有四个边缘都被水包围。
假设网格就像-
有三个岛屿。
为了解决这个问题，我们将遵循以下步骤-
将有两种方法，一种将用于计算称为numIslands()和的岛的数量makeWater()。该makeWater()会像-
如果网格中的行数为0，则返回0
n=行数，m：=列数，ans：=0
对于i，范围为0至n–1
如果grid[i，j]=1，则ans：=ans+1
makeWater(i,j,n,m,grid)
对于0到m范围内的j
makeWater()将采取指数I，J，ROW和COL计数n和m和网格
如果i<0或j<0或i>=n或j>=m，则从此方法返回
如果grid[i，j]=0，则返回，否则返回grid[i，j]：=0
调用makeWater（i+1，j，n，m，grid）
调用makeWater（i，j+1，n，m，grid）
让我们看下面的实现以更好地理解-
示例
class Solution(object):
def numIslands(self, grid):
if len(grid) == 0:
return 0
n= len(grid)
m = len(grid[0])
ans = 0
for i in range(n):
for j in range(m):
if grid[i][j] == "1":
ans+=1
self.make_water(i,j,n,m,grid)
return ans
def make_water(self,i,j,n,m,grid):
if i<0 or j<0 or i>=n or j>=m:
return
if grid[i][j] == "0":
return
else:
grid[i][j]="0"
self.make_water(i+1,j,n,m,grid)
self.make_water(i,j+1,n,m,grid)
self.make_water(i-1,j,n,m,grid)
self.make_water(i,j-1,n,m,grid)
ob1 = Solution()print(ob1.numIslands([["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],
["0","0","0","1","1"]]))输入项
[["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],["0","0","0","1","1"]]
输出结果
3