feat: add solutions to lc problem: No.0309. Best Time to Buy and Sell Stock with Cooldown

Author: yanglbme

README.md

@@ -174,6 +174,7 @@
 - [最佳观光组合](./solution/1000-1099/1014.Best%20Sightseeing%20Pair/README.md)
 - [买卖股票的最佳时机](./solution/0100-0199/0121.Best%20Time%20to%20Buy%20and%20Sell%20Stock/README.md)
 - [买卖股票的最佳时机 II](./solution/0100-0199/0122.Best%20Time%20to%20Buy%20and%20Sell%20Stock%20II/README.md)
+- [最佳买卖股票时机含冷冻期](./solution/0300-0399/0309.Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Cooldown/README.md)
 - [接雨水](./solution/0000-0099/0042.Trapping%20Rain%20Water/README.md)
 - [礼物的最大价值](./lcof/面试题47.%20礼物的最大价值/README.md)
 - [最小路径和](./solution/0000-0099/0064.Minimum%20Path%20Sum/README.md)

README_EN.md

@@ -168,6 +168,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
 - [Best Sightseeing Pair](./solution/1000-1099/1014.Best%20Sightseeing%20Pair/README_EN.md)
 - [Best Time to Buy and Sell Stock](./solution/0100-0199/0121.Best%20Time%20to%20Buy%20and%20Sell%20Stock/README_EN.md)
 - [Best Time to Buy and Sell Stock II](./solution/0100-0199/0122.Best%20Time%20to%20Buy%20and%20Sell%20Stock%20II/README_EN.md)
+- [Best Time to Buy and Sell Stock with Cooldown](./solution/0300-0399/0309.Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Cooldown/README_EN.md)
 - [Trapping Rain Water](./solution/0000-0099/0042.Trapping%20Rain%20Water/README_EN.md)
 - [Minimum Path Sum](./solution/0000-0099/0064.Minimum%20Path%20Sum/README_EN.md)
 - [Decode Ways](./solution/0000-0099/0091.Decode%20Ways/README_EN.md)

solution/0300-0399/0309.Best Time to Buy and Sell Stock with Cooldown/README.md

@@ -21,27 +21,108 @@
 <strong>输出: </strong>3 
 <strong>解释:</strong> 对应的交易状态为: [买入, 卖出, 冷冻期, 买入, 卖出]</pre>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+动态规划法。
+
+设 f1 表示当天买入股票后的最大利润，f2 表示当天卖出股票后的最大利润，f3 表示当天空仓后的最大利润。
+
+初始第 1 天结束时，`f1 = -prices[0]`，`f2 = 0`，`f3 = 0`。
+
+从第 2 天开始，当天结束时：
+
+- 若买入，则说明前一天空仓，然后今天买入，`f1 = max(f1, f3 - prices[i])`。
+- 若卖出，则只能是之前某一天买入，然后今天卖出，`f2 = max(f2, f1 + prices[i])`。
+- 若空仓，则只能是之前某一天卖出后，然后今天保持空仓，`f3 = max(f3, f2)`。
+
+最后返回 f2 即可。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        # 买入，卖出，继续空仓
+        f1, f2, f3 = -prices[0], 0, 0
+        res = f2
+        for price in prices[1:]:
+            pf1, pf2, pf3 = f1, f2, f3
+            f1 = max(pf1, pf3 - price)
+            f2 = max(pf2, pf1 + price)
+            f3 = max(pf3, pf2)
+            res = max(res, f2)
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int maxProfit(int[] prices) {
+        // 买入，卖出，继续空仓
+        int f1 = -prices[0], f2 = 0, f3 = 0;
+        int res = f2;
+        for (int i = 1; i < prices.length; ++i) {
+            int pf1 = f1, pf2 = f2, pf3 = f3;
+            f1 = Math.max(pf1, pf3 - prices[i]);
+            f2 = Math.max(pf2, pf1 + prices[i]);
+            f3 = Math.max(pf3, pf2);
+            res = Math.max(res, f2);
+        }
+        return res;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        int f1 = -prices[0], f2 = 0, f3 = 0;
+        int res = f2;
+        for (int i = 1; i < prices.size(); ++i) {
+            int pf1 = f1, pf2 = f2, pf3 = f3;
+            f1 = max(pf1, pf3 - prices[i]);
+            f2 = max(pf2, pf1 + prices[i]);
+            f3 = max(pf3, pf2);
+            res = max(res, f2);
+        }
+        return res;
+    }
+};
+```
 
+### **Go**
+
+```go
+func maxProfit(prices []int) int {
+	f1, f2, f3, res := -prices[0], 0, 0, 0
+	for i := 1; i < len(prices); i++ {
+		pf1, pf2, pf3 := f1, f2, f3
+		f1 = max(pf1, pf3-prices[i])
+		f2 = max(pf2, pf1+prices[i])
+		f3 = max(pf3, pf2)
+		res = max(res, f2)
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
 ```
 
 ### **...**

solution/0300-0399/0309.Best Time to Buy and Sell Stock with Cooldown/README_EN.md

@@ -46,13 +46,79 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        f1, f2, f3 = -prices[0], 0, 0
+        res = f2
+        for price in prices[1:]:
+            pf1, pf2, pf3 = f1, f2, f3
+            f1 = max(pf1, pf3 - price)
+            f2 = max(pf2, pf1 + price)
+            f3 = max(pf3, pf2)
+            res = max(res, f2)
+        return res
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public int maxProfit(int[] prices) {
+        int f1 = -prices[0], f2 = 0, f3 = 0;
+        int res = f2;
+        for (int i = 1; i < prices.length; ++i) {
+            int pf1 = f1, pf2 = f2, pf3 = f3;
+            f1 = Math.max(pf1, pf3 - prices[i]);
+            f2 = Math.max(pf2, pf1 + prices[i]);
+            f3 = Math.max(pf3, pf2);
+            res = Math.max(res, f2);
+        }
+        return res;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        int f1 = -prices[0], f2 = 0, f3 = 0;
+        int res = f2;
+        for (int i = 1; i < prices.size(); ++i) {
+            int pf1 = f1, pf2 = f2, pf3 = f3;
+            f1 = max(pf1, pf3 - prices[i]);
+            f2 = max(pf2, pf1 + prices[i]);
+            f3 = max(pf3, pf2);
+            res = max(res, f2);
+        }
+        return res;
+    }
+};
+```
 
+### **Go**
+
+```go
+func maxProfit(prices []int) int {
+	f1, f2, f3, res := -prices[0], 0, 0, 0
+	for i := 1; i < len(prices); i++ {
+		pf1, pf2, pf3 := f1, f2, f3
+		f1 = max(pf1, pf3-prices[i])
+		f2 = max(pf2, pf1+prices[i])
+		f3 = max(pf3, pf2)
+		res = max(res, f2)
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
 ```
 
 ### **...**

solution/0300-0399/0309.Best Time to Buy and Sell Stock with Cooldown/Solution.cpp

@@ -0,0 +1,15 @@
+class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        int f1 = -prices[0], f2 = 0, f3 = 0;
+        int res = f2;
+        for (int i = 1; i < prices.size(); ++i) {
+            int pf1 = f1, pf2 = f2, pf3 = f3;
+            f1 = max(pf1, pf3 - prices[i]);
+            f2 = max(pf2, pf1 + prices[i]);
+            f3 = max(pf3, pf2);
+            res = max(res, f2);
+        }
+        return res;
+    }
+};

solution/0300-0399/0309.Best Time to Buy and Sell Stock with Cooldown/Solution.go

@@ -0,0 +1,18 @@
+func maxProfit(prices []int) int {
+	f1, f2, f3, res := -prices[0], 0, 0, 0
+	for i := 1; i < len(prices); i++ {
+		pf1, pf2, pf3 := f1, f2, f3
+		f1 = max(pf1, pf3-prices[i])
+		f2 = max(pf2, pf1+prices[i])
+		f3 = max(pf3, pf2)
+		res = max(res, f2)
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}

solution/0300-0399/0309.Best Time to Buy and Sell Stock with Cooldown/Solution.java

@@ -1,20 +1,14 @@
 class Solution {
     public int maxProfit(int[] prices) {
-        if(prices == null || prices.length == 0) return 0;
-        
-        int n = prices.length;
-        // buy[i] 表示第i天持有股票，最大利润
-        int[] buy = new int[n];
-        // sell[i] 表示第i天为持股，最大利润
-        int[] sell = new int[n];
-        buy[0] = -prices[0];
-        sell[0] = 0;
-        
-        for(int i = 1; i < n; i++) {
-            buy[i] = Math.max(buy[i - 1], (i > 1 ? sell[i - 2] : 0) - prices[i]);
-            sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]);
+        int f1 = -prices[0], f2 = 0, f3 = 0;
+        int res = f2;
+        for (int i = 1; i < prices.length; ++i) {
+            int pf1 = f1, pf2 = f2, pf3 = f3;
+            f1 = Math.max(pf1, pf3 - prices[i]);
+            f2 = Math.max(pf2, pf1 + prices[i]);
+            f3 = Math.max(pf3, pf2);
+            res = Math.max(res, f2);
         }
-        
-        return sell[n - 1];
+        return res;
     }
 }

solution/0300-0399/0309.Best Time to Buy and Sell Stock with Cooldown/Solution.py

@@ -0,0 +1,11 @@
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        f1, f2, f3 = -prices[0], 0, 0
+        res = f2
+        for price in prices[1:]:
+            pf1, pf2, pf3 = f1, f2, f3
+            f1 = max(pf1, pf3 - price)
+            f2 = max(pf2, pf1 + price)
+            f3 = max(pf3, pf2)
+            res = max(res, f2)
+        return res