算法训练营Day50（动态规划11）

提醒

这道题一下子就难度上来了，关键在于至多买卖两次，这意味着可以买卖一次，可以买卖两次，也可以不买卖

class Solution:def maxProfit(self, prices: List[int]) -> int:if len(prices) == 0:return 0dp = [[0] * 5 for _ in range(len(prices))]dp[0][1] = -prices[0]dp[0][3] = -prices[0]for i in range(1, len(prices)):dp[i][0] = dp[i-1][0]dp[i][1] = max(dp[i-1][1], dp[i-1][0] - prices[i])dp[i][2] = max(dp[i-1][2], dp[i-1][1] + prices[i])dp[i][3] = max(dp[i-1][3], dp[i-1][2] - prices[i])dp[i][4] = max(dp[i-1][4], dp[i-1][3] + prices[i])return dp[-1][4]

188.买卖股票的最佳时机IV  力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

提醒

本题是123.买卖股票的最佳时机III 的进阶版  

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