Leetcode 188 买卖股票的最佳时机IV
class Solution:def maxProfit(self, k: int, prices: List[int]) -> int:if not prices: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]
leetcode309 买卖股票的最佳时机涵冷冻期
class Solution:def maxProfit(self, prices: List[int]) -> int:n = len(prices)dp = [[0] * 4 for i in range(n)]dp[0][0] = -prices[0]for i in range(1, n):dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][1], dp[i - 1][3]) - prices[i]) # 持有股票dp[i][1] = max(dp[i - 1][1], dp[i - 1][3]) # 不持有股票且处于冷静期dp[i][2] = dp[i - 1][0] + prices[i] # 不持有股票即卖出dp[i][3] = dp[i - 1][2] # 处于冷静期return max(dp[n-1][3], dp[n-1][2], dp[n-1][1])
leetcode 714 买卖股票的最佳时机含手续费
class Solution:def maxProfit(self, prices: List[int], fee: int) -> int:n = len(prices)dp = [[0] * 2 for _ in range(n)]dp[0][0] = -prices[0]for i in range(1, n):dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i])dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i] - fee)return max(dp[-1][0], dp[-1][1])
