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和/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:grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
输出:1
示例 2:
输入:grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出:3
提示:
m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/number-of-islands
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution: def numIslands(self, grid: List[List[str]]) -> int: m,n=len(grid), len(grid[0]) visited=[[0 for i in range(n)] for j in range(m)] res=0 for i in range(m): for j in range(n): if visited[i][j]==0 and grid[i][j]=='1': res+=1 self.dfs(m, n, i, j, grid, visited) return res def dfs(self, m, n, i, j, grid, visited): tmp=[[i,j]] visited[i][j]=1 while tmp: i,j=tmp.pop() if i+1<m and visited[i+1][j]==0 and grid[i+1][j]=='1': visited[i+1][j]=1 tmp.append([i+1, j]) if j+1<n and visited[i][j+1]==0 and grid[i][j+1]=='1': visited[i][j+1]=1 tmp.append([i, j+1]) if i-1>=0 and visited[i-1][j]==0 and grid[i-1][j]=='1': visited[i-1][j]=1 tmp.append([i-1, j]) if j-1>=0 and visited[i][j-1]==0 and grid[i][j-1]=='1': visited[i][j-1]=1 tmp.append([i, j-1])