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73 lines
1.6 KiB
Python
73 lines
1.6 KiB
Python
"""
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200. Number of Islands
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Difficulty: Medium
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https://leetcode.com/problems/number-of-islands/
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──────────────────────────────────────────────────
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Given an m x n 2D binary grid grid which represents a map of '1's
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(land) and '0's (water), return the number of islands.
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An island is surrounded by water and is formed by connecting adjacent
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lands horizontally or vertically. You may assume all four edges of the
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grid are all surrounded by water.
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Example 1:
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Input: grid = [
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["1","1","1","1","0"],
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["1","1","0","1","0"],
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["1","1","0","0","0"],
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["0","0","0","0","0"]
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]
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Output: 1
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Example 2:
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Input: grid = [
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["1","1","0","0","0"],
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["1","1","0","0","0"],
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["0","0","1","0","0"],
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["0","0","0","1","1"]
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]
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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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"""
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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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