200. Number of Islands

[fockspaces]

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'.

[fockspaces]

用 recursion travel through all the islands，並將走過的 islands 變成海

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        m, n, count = len(grid), len(grid[0]), 0
        def DFS(grid, r, c):
            if r < 0 or c < 0 or r >= m or c >= n:
                return 
            if grid[r][c] != '1':
                return
            grid[r][c] = '0'
            DFS(grid, r + 1, c)
            DFS(grid, r - 1, c)
            DFS(grid, r, c + 1)
            DFS(grid, r, c - 1)
        for row in range(m):
            for col in range(n):
                if grid[row][col] == '1':
                    DFS(grid, row, col)
                    count += 1
        return count

[fockspaces]

可以更進一步將上下左右用 for loop 表示

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        m, n, count = len(grid), len(grid[0]), 0
        directions = [[1, 0], [-1, 0], [0, 1], [0, -1]]
        def DFS(grid, r, c):
            if r < 0 or c < 0 or r >= m or c >= n:
                return 
            if grid[r][c] != '1':
                return
            grid[r][c] = '0'
            for dir in directions:
                DFS(grid, r + dir[0], c + dir[1])
        for row in range(m):
            for col in range(n):
                if grid[row][col] == '1':
                    DFS(grid, row, col)
                    count += 1
        return count