Create 200_Number_of_Islands.md

Author: koji

200_Number_of_Islands.md

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+## 200. Number of Islands
+
+Given an m x n 2d grid 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
+def numIslands(self, grid: List[List[str]]) -> int:
+ r = len(grid)
+ c = len(grid[0])
+ 
+ count = 0
+ 
+ def dfs(grid, i, j):
+
+
+ if i <0 or j <0 or i>=r or j>=c or grid[i][j] != '1':
+ return
+
+ grid[i][j] = '#'
+
+ dfs(grid, i+1, j)
+ dfs(grid, i-1, j)
+ dfs(grid, i, j+1)
+ dfs(grid, i, j-1)
+ 
+ for i in range(r):
+ for j in range(c):
+ if grid[i][j] == '1':
+ dfs(grid, i, j)
+ count +=1
+ return count
+```
+
+
+
+```
+Runtime: 132 ms, faster than 86.94% of Python3 online submissions for Number of Islands.
+Memory Usage: 15.5 MB, less than 38.11% of Python3 online submissions for Number of Islands.
+```