遍历二维数组,找到为 1 的之后,使用深度优先搜索将与其相邻的陆地全部淹没(也就是1 改为 0)
#
# 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
#
class Solution:
def solve(self , grid: List[List[str]]) -> int:
# write code here
m=len(grid)
n=len(grid[0])
num=0
def dfs(grid, x, y):
# 先“淹掉”自己
grid[x][y] = '0'
if x + 1 < m and grid[x+1][y] == '1': dfs(grid,x+1, y)
if x - 1 >= 0 and grid[x-1][y] == '1': dfs(grid,x-1, y)
if y + 1 < n and grid[x][y+1] == '1': dfs(grid,x, y+1)
if y - 1 >= 0 and grid[x][y-1] == '1': dfs(grid,x, y-1)
for i in range(m):
for j in range(n):
if grid[i][j]=='1':
dfs(grid,i,j)
num+=1
return num

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