200. Number of Islands

200_number_of_islands/problem.md

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+## 問題: [200. Number of Islands](https://leetcode.com/problems/number-of-islands/description/)

200_number_of_islands/step1.md

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+# Step 1
+
+- 各セルで条件を満たす場合、dfsを行う
+  - dfs
+    - セルが1、まだ訪れていない、かつ範囲内だったら、counterを1増やして上下左右に行き、次のセルでも同じことを繰り返す
+
+```python
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        rows = len(grid)
+        columns = len(grid[0])
+        visited = set()
+        num_islands = 0
+
+        def dfs(r, c):
+            if (r, c) not in visited and r >= 0 and r < rows and c >= 0 and c < columns and grid[r][c] == "1":
+                visited.add((r, c))
+                dfs(r + 1, c)
+                dfs(r - 1, c)
+                dfs(r, c + 1)
+                dfs(r, c - 1)
+
+        for r in range(rows):
+            for c in range(columns):
+                if grid[r][c] == "1" and (r, c) not in visited:
+                    num_islands += 1
+                    dfs(r, c)
+
+        return num_islands
+```
+
+時間計算量: $O(mn)$
+
+空間計算量: $O(mn)$
+
+アルゴリズム最大実行時間(概算): 時間計算量 / Pythonの1秒あたりの計算ステップ数 -> $(300 \times 300) \div 10,000,000 = 0.001 (s)$

200_number_of_islands/step2.md

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+# Step 2
+
+```python
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        rows = len(grid)
+        columns = len(grid[0])
+        directions = [[1, 0], [-1, 0], [0, 1], [0, -1]]
+        visited = set()
+        num_islands = 0
+
+        def dfs(r, c):
+            if not (0 <= r < rows and 0 <= c < columns):
+                return
+            if grid[r][c] != "1" or (r, c) in visited:
+                return
+            visited.add((r, c))
+            for dr, dc in directions:
+                dfs(r + dr, c + dc)
+
+        for r in range(rows):
+            for c in range(columns):
+                if grid[r][c] == "1" and (r, c) not in visited:
+                    num_islands += 1
+                    dfs(r, c)
+
+        return num_islands
+```
+
+時間計算量: $O(mn)$
+
+空間計算量: $O(mn)$
+
+アルゴリズム最大実行時間(概算): 時間計算量 / Pythonの1秒あたりの計算ステップ数 -> $(300 \times 300) \div 10,000,000 = 0.001 (s)$

200_number_of_islands/step3.md

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+# Step 3
+
+```python
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        rows = len(grid)
+        columns = len(grid[0])
+        directions = [[1, 0], [-1, 0], [0, 1], [0, -1]]
+        visited = set()
+        num_islands = 0
+
+        def dfs(r, c):
+            if not (0 <= r < rows and 0 <= c < columns):
+                return
+            if grid[r][c] != "1" or (r, c) in visited:
+                return
+            visited.add((r, c))
+            for dr, dc in directions:
+                dfs(r + dr, c + dc)
+
+        for r in range(rows):
+            for c in range(columns):
+                if grid[r][c] == "1" and (r, c) not in visited:
+                    num_islands += 1
+                    dfs(r, c)
+
+        return num_islands
+```
+1回目: 5分　9秒
+
+2回目: 3分　25秒
+
+3回目: 2分　57秒