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57 lines
1.6 KiB
Plaintext
57 lines
1.6 KiB
Plaintext
---
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title: '200. Number of Islands'
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description: Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands
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icon: dot
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topics:
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- { name: "Array", slug: "array" }
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- { name: "Depth-First Search", slug: "depth-first-search" }
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- { name: "Breadth-First Search", slug: "breadth-first-search" }
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- { name: "Union-Find", slug: "union-find" }
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- { name: "Matrix", slug: "matrix" }
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sidebar:
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label: 'Number of Islands'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `grid = [ ["1","1","1","1","0"], ["1","1","0","1","0"], ["1","1","0","0","0"], ["0","0","0","0","0"] ]`
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- Output: `1`
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### Example 2:
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- Input: `grid = [ ["1","1","0","0","0"], ["1","1","0","0","0"], ["0","0","1","0","0"], ["0","0","0","1","1"] ]`
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- Output: `3`
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### Constraints:
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- `m == grid.length`
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- `n == grid[i].length`
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- `1 <= m, n <= 300`
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- grid[i][j] is '0' or '1'.
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## Solution
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```py
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class Solution:
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def numIslands(self, grid: List[List[str]]) -> int:
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rows, cols = len(grid), len(grid[0])
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def dfs(r, c):
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if r < 0 or r >= rows or c < 0 or c >= cols or grid[r][c] != "1":
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return
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else:
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grid[r][c] = "0"
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dfs(r, c + 1) # right
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dfs(r + 1, c) # bottom
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dfs(r, c - 1) # left
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dfs(r - 1, c) # top
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num_islands = 0
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for r in range(rows):
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for c in range(cols):
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if grid[r][c] == "1":
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num_islands += 1
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dfs(r, c)
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return num_islands
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```
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