[leetcode] 200.Number of Islands

200. Number of Islands
     Medium
     Topics
     Companies
     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

grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]

rows, cols = len(grid), len(grid[0])
print('rows:', rows, 'cols:', cols)

def dfs(i,j):
  #더 이상 땅이 아닌 경우 종료
  if i < 0 or i >= rows or j < 0 or j >= cols or grid[i][j] != '1':
    return

  grid[i][j] = '0'
  #동서남북 탐색
  dfs(i+1, j)
  dfs(i-1, j)
  dfs(i, j+1)
  dfs(i, j-1)

count = 0
for i in range(rows):
  for j in range(cols):
    if grid[i][j] == '1':
      dfs(i,j)
      count += 1
print(count)