Merge pull request #1151 from berserk-112/master

Update 0309.最佳买卖股票时机含冷冻期.md
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程序员Carl
2022-03-28 09:55:50 +08:00
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2 changed files with 23 additions and 1 deletions

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@ -1,5 +1,4 @@
👉 推荐 [在线阅读](http://programmercarl.com/) (Github在国内访问经常不稳定) 👉 推荐 [在线阅读](http://programmercarl.com/) (Github在国内访问经常不稳定)
👉 推荐 [Gitee同步](https://gitee.com/programmercarl/leetcode-master) 👉 推荐 [Gitee同步](https://gitee.com/programmercarl/leetcode-master)

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@ -205,6 +205,29 @@ class Solution {
} }
} }
``` ```
```java
//另一种解题思路
class Solution {
public int maxProfit(int[] prices) {
int[][] dp = new int[prices.length + 1][2];
dp[1][0] = -prices[0];
for (int i = 2; i <= prices.length; i++) {
/*
dp[i][0] 第i天未持有股票收益;
dp[i][1] 第i天持有股票收益;
情况一第i天是冷静期不能以dp[i-1][1]购买股票,所以以dp[i - 2][1]买股票,没问题
情况二第i天不是冷静期理论上应该以dp[i-1][1]购买股票但是第i天不是冷静期说明第i-1天没有卖出股票
则dp[i-1][1]=dp[i-2][1],所以可以用dp[i-2][1]买股票,没问题
*/
dp[i][0] = Math.max(dp[i - 1][0], dp[i - 2][1] - prices[i - 1]);
dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] + prices[i - 1]);
}
return dp[prices.length][1];
}
}
```
Python Python