[Leetcode] 0200. Number Of Islands

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內容目錄

Medium

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.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2:

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] is '0' or '1'.

Python

				
					# time complexity: O(m*n)
# space complexity: O(m*n)
from typing import List


class Solution:
    def dfs(self, grid: List[List[str]], r: int, c: int) -> None:
        nR = len(grid)
        nC = len(grid[0])
        grid[r][c] = '0'
        if r - 1 >= 0 and grid[r - 1][c] == "1":
            self.dfs(grid, r - 1, c)
        if r + 1 < nR and grid[r + 1][c] == "1":
            self.dfs(grid, r + 1, c)
        if c - 1 >= 0 and grid[r][c - 1] == "1":
            self.dfs(grid, r, c - 1)
        if c + 1 < nC and grid[r][c + 1] == "1":
            self.dfs(grid, r, c + 1)

    def numIslands(self, grid: List[List[str]]) -> int:
        if not grid:
            return 0
        numIslands = 0
        for r in range(len(grid)):
            for c in range(len(grid[0])):
                if grid[r][c] == '1':
                    numIslands += 1
                    self.dfs(grid, r, c)
        return numIslands


grid = [
    ["1", "1", "0", "0", "0"],
    ["1", "1", "0", "0", "0"],
    ["0", "0", "1", "0", "0"],
    ["0", "0", "0", "1", "1"]
]
print(Solution().numIslands(grid))

Medium

Python

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