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