Description
Given a 2d grid map of '1's (land) and '0's (water), count 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:
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Input:
11110
11010
11000
00000
Output: 1
Example 2:
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Input:
11000
11000
00100
00011
Output: 3
Approach 1
A recursive way of checking every element and its four neighbors of the grid to see if it is ‘1’ and mark as ‘0’ if so.
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class Solution:
def isValid(self, grid, x, y):
if x < 0 or y < 0 or x >= len(grid) or y >= len(grid[0]):
return False
return True
def dfs(self, grid, x, y):
grid[x][y] = "0"
neighbour_vec = [(0, -1), (0, 1), (-1, 0), (1, 0)]
for neighbour in neighbour_vec:
neigh_x = x + neighbour[0]
neigh_y = y + neighbour[1]
if self.isValid(grid, neigh_x, neigh_y) and grid[neigh_x][neigh_y] == "1":
self.dfs(grid, neigh_x, neigh_y)
def numIslands(self, grid: List[List[str]]) -> int:
count = 0
if not grid:
return 0
x_dim = len(grid[0])
y_dim = len(grid)
for i in range(y_dim):
for j in range(x_dim):
if grid[i][j] == "1":
self.dfs(grid, i, j)
count += 1
return count
