Leetcode刷题笔记1 动态规划part09

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 0
        dp = [[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])