122. Best Time to Buy and Sell Stock II

122_best_time_to_buy_and_sell_stock_ii/problem.md

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+## 問題: [122. Best Time to Buy and Sell Stock II](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/)

122_best_time_to_buy_and_sell_stock_ii/step1.md

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+# Step 1
+
+- 各整数を折れ線グラフの点と考え、右上がりになるものを合計に加えていく
+- 上記を実現する式: $maxProfit=maxProfit + max(prices[i] - prices[i-1], 0)$
+
+```python
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        if len(prices) <= 1:
+            return 0
+        max_profit = 0
+        for i in range(1, len(prices)):
+            max_profit += max(prices[i] - prices[i - 1], 0)
+        return max_profit
+```
+
+時間計算量: $O(n)$
+
+空間計算量: $O(1)$

122_best_time_to_buy_and_sell_stock_ii/step2.md

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+# Step 2
+
+```python
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        if len(prices) < 2:
+            return 0
+        max_profit = 0
+        for i in range(1, len(prices)):
+            max_profit += max(prices[i] - prices[i - 1], 0)
+        return max_profit
+```
+
+時間計算量: $O(n)$
+
+空間計算量: $O(1)$

122_best_time_to_buy_and_sell_stock_ii/step3.md

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+# Step 3
+
+```python
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        if len(prices) < 2:
+            return 0
+        max_profit = 0
+        for i in range(1, len(prices)):
+            max_profit += max(prices[i] - prices[i - 1], 0)
+        return max_profit
+```
+1回目: 0分　56秒
+
+2回目: 0分　48秒
+
+3回目: 0　53秒