Leetcode 200. number of Islands

Posted 2020-08-19 lettuan

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:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

 

本题使用DFS或者BFS都可以。

Solution. 1 DFS.

 1 class Solution(object):
 2 
 3     def numIslands(self, grid):
 4         """
 5         :type grid: List[List[str]]
 6         :rtype: int
 7         """
 8         m = len(grid)
 9         if m == 0:
10             return 0
11         n = len(grid[0])
12         
13         ans = 0
14         for i in range(m):
15             for j in range(n):
16                 if grid[i][j] == ‘1‘:
17                     ans += 1
18                     self.helper(grid, m, n, i, j)
19                 
20         return ans
21         
22     def helper(self, grid, m, n, i, j):
23         if (i < 0 or j < 0 or i > m-1 or j > n-1):
24             return 
25         
26         if grid[i][j] == ‘1‘:
27             grid[i][j] = ‘2‘
28             self.helper(grid, m, n, i-1, j)
29             self.helper(grid, m, n, i+1, j )
30             self.helper(grid, m, n, i, j-1)
31             self.helper(grid, m, n, i, j+1)

Solution. 2 BFS

 1 class Solution(object):
 2     def numIslands(self, grid):
 3         """
 4         :type grid: List[List[str]]
 5         :rtype: int
 6         """
 7         m = len(grid)
 8         if m == 0:
 9             return 0
10         n = len(grid[0])
11         visited = [ [False]* n for x in range(m)]
12         ans = 0
13         for i in range(m):
14             for j in range(n):
15                 if grid[i][j] == ‘1‘ and visited[i][j] == False:
16                     ans += 1
17                     self.bfs(grid, visited, m, n, i, j)
18                 
19         return ans
20         
21     def bfs(self, grid, visited, m, n, i, j):
22         direct = [[0,1],[0,-1],[1,0],[-1,0]]
23         queue = [[i, j]]
24         visited[i][j] = True
25         
26         while queue:
27             front = queue.pop(0)
28             for p in direct:
29                 np = [front[0]+p[0], front[1]+p[1]]
30                 if self.isValid(np, m, n) and grid[np[0]][np[1]] == ‘1‘31                 and visited[np[0]][np[1]] == False:
32                     visited[np[0]][np[1]] = True
33                     queue.append(np)
34     
35     def isValid(self, np, m, n):
36         return np[0]>=0 and np[0]<m and np[1]>=0 and np[1]<n