代码随性训练营第五十天（Python）｜123.买卖股票的最佳时机III 、188.买卖股票的最佳时机IV

123.买卖股票的最佳时机III

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if len(prices) <= 1:
            return 0
        if len(prices) == 2:
            return max(0, prices[1] - prices[0])
        if len(prices) == 3:
            return max(prices[2] - prices[1], prices[1]- prices[0], prices[2]- prices[0], 0)
        # dp[i][0] 第 i 天不操作
        # dp[i][1] 第 i 天第一次持有
        # dp[i][2] 第 i 天第一次卖出
        # dp[i][3] 第 i 天第二次持有
        # dp[i][4] 第 i 天第二次卖出
        dp = [[0] * 5 for _ in range(len(prices))]
        dp[0][0] = 0
        dp[0][1] = -prices[0]
        dp[0][2] = 0
        dp[0][3] = -prices[0]
        dp[0][4] = 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

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