和前几个一样，dp数组变大到[n][2k+1]，推导公式也一样。

class Solution:def maxProfit(self, k: int, prices: List[int]) -> int:dp = [[0]*(2*k+1) for _ in range(len(prices))]for i in range(k):dp[0][2*i+1] = -prices[0]for i in range(1, len(prices)):for j in range(1, 2*k+1):if j % 2 == 0:dp[i][j] = max(dp[i-1][j], dp[i-1][j-1]+prices[i])else:dp[i][j] = max(dp[i-1][j], dp[i-1][j-1]-prices[i])return dp[-1][-1]

dp中多一个冷静期，别的还是一样的。

class Solution:def maxProfit(self, prices: List[int]) -> int:dp = [[0]*3 for _ in prices]dp[0][0] = -prices[0]for i in range(1, len(prices)):dp[i][0] = max(dp[i-1][0], dp[i-1][2]-prices[i])dp[i][1] = max(dp[i-1][1], dp[i-1][0]+prices[i])dp[i][2] = dp[i-1][1]return dp[-1][1]
```# [714. 买卖股票的最佳时机含手续费](https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/)
买入的时候多扣除一个买卖费用就行了，别的也一样。
```python
class Solution:def maxProfit(self, prices: List[int], fee: int) -> int:dp = [[0] * 2 for _ in prices]dp[0][0] = -prices[0] - feefor i in range(1, len(prices)):dp[i][0] = max(dp[i-1][0], dp[i-1][1]-prices[i]-fee)dp[i][1] = max(dp[i-1][0]+prices[i], dp[i-1][1])return dp[-1][-1]
```