From 127986e03a0b4acdedf80a6b38d3ded7d61514f3 Mon Sep 17 00:00:00 2001
From: Steve2020 <841532108@qq.com>
Date: Fri, 13 May 2022 16:19:59 +0800
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---
 ...09.最佳买卖股票时机含冷冻期.md | 60 +++++++++++++++++++
 1 file changed, 60 insertions(+)

diff --git a/problems/0309.最佳买卖股票时机含冷冻期.md b/problems/0309.最佳买卖股票时机含冷冻期.md
index f3e7541b..f037fe85 100644
--- a/problems/0309.最佳买卖股票时机含冷冻期.md
+++ b/problems/0309.最佳买卖股票时机含冷冻期.md
@@ -325,6 +325,66 @@ const maxProfit = (prices) => {
 };
 ```
 
+TypeScript：
+
+> 版本一，与本文思路一致
+
+```typescript
+function maxProfit(prices: number[]): number {
+    /**
+        dp[i][0]: 持股状态；
+        dp[i][1]: 无股状态，当天为非冷冻期；
+        dp[i][2]: 无股状态，当天卖出；
+        dp[i][3]: 无股状态，当天为冷冻期；
+     */
+    const length: number = prices.length;
+    const dp: number[][] = new Array(length).fill(0).map(_ => []);
+    dp[0][0] = -prices[0];
+    dp[0][1] = dp[0][2] = dp[0][3] = 0;
+    for (let i = 1; i < length; i++) {
+        dp[i][0] = Math.max(
+            dp[i - 1][0],
+            Math.max(dp[i - 1][1], dp[i - 1][3]) - prices[i]
+        );
+        dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][3]);
+        dp[i][2] = dp[i - 1][0] + prices[i];
+        dp[i][3] = dp[i - 1][2];
+    }
+    const lastEl: number[] = dp[length - 1];
+    return Math.max(lastEl[1], lastEl[2], lastEl[3]);
+};
+```
+
+> 版本二，状态定义略有不同，可以帮助理解
+
+```typescript
+function maxProfit(prices: number[]): number {
+    /**
+        dp[i][0]: 持股状态，当天买入；
+        dp[i][1]: 持股状态，当天未买入；
+        dp[i][2]: 无股状态，当天卖出；
+        dp[i][3]: 无股状态，当天未卖出；
+        
+        买入有冷冻期限制，其实就是状态[0]只能由前一天的状态[3]得到；
+        如果卖出有冷冻期限制，其实就是[2]由[1]得到。
+     */
+    const length: number = prices.length;
+    const dp: number[][] = new Array(length).fill(0).map(_ => []);
+    dp[0][0] = -prices[0];
+    dp[0][1] = -Infinity;
+    dp[0][2] = dp[0][3] = 0;
+    for (let i = 1; i < length; i++) {
+        dp[i][0] = dp[i - 1][3] - prices[i];
+        dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0]);
+        dp[i][2] = Math.max(dp[i - 1][0], dp[i - 1][1]) + prices[i];
+        dp[i][3] = Math.max(dp[i - 1][3], dp[i - 1][2]);
+    }
+    return Math.max(dp[length - 1][2], dp[length - 1][3]);
+};
+```
+
+
+
 
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