leetcode动态规划—买卖股票系列

lc121 买卖股票的最佳时机

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        #dp[i][0] 第i天不持有股票，最大利润
        #dp[i][1] 第i天持有股票，最大利润
        #dp[i][0] = max(dp[i-1][0], dp[i-1][1] + prices[i])
        #dp[i][1] = max(dp[i-1][1], -prices[i])
        
        n = len(prices)
        #初始化
        dp = [[0] * 2 for _ in range(n)]
        dp[0][0] = 0
        dp[0][1] = -prices[0]

        #遍历
        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], -prices[i])
        return dp[n-1][0]

lc122 买卖股票的最佳时机II

可以多次买卖，买股票时，手头的现金不再是0，而是前一天的dp[i-1][0]

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        #dp[i][0]: 第i天不持有股票，最大利润
        #dp[i][1]: 第i天持有股票，最大利润
        #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])
        n = len(prices)
        dp = [[0] * 2 for _ in range(n)]
        dp[0][0] = 0
        dp[0][1] = -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])

        return dp[n-1][0]

lc123 买卖股票的最佳时机III

多定义几个状态

2次交易

2*2 + 1 个状态

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        #dp[i][0]:第i天，第0次不持有，最大利润
        #dp[i][1]:第i天，第1次持有，最大利润
        #dp[i][2]:第i天，第1次不持有，最大利润
        #dp[i][3]:第i天，第2次持有，最大利润
        #dp[i][4]:第i天，第2次不持有，最大利润
    
        n = len(prices)
        dp = [[0] * 5 for _ in range(n)]
        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, n):
            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[n-1][4]

lc188 买卖股票的最佳时机IV

把上述的2次交易，换成k次

k次交易

2*k + 1个状态

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

        for i in range(1, n):
            dp[i][0] = dp[i-1][0]
            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[n-1][2 * k]

lc309 买卖股票的最佳时机+冷冻期

将不持有状态分为 卖出 \ 非卖出

对于第i天持有：

        要么从 第i-1天 持有继承

        要么从 第i-1天 ”不持有且非卖出状态“ 买入

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        #dp[i][0] : 第i天不持有且非卖出, 最大利润
        #dp[i][1] : 第i天不持有且卖出，最大利润
        #dp[i][2] : 第i天持有，最大利润

        #dp[i][0] = max(dp[i-1][0], dp[i-1][1])
        #dp[i][1] = dp[i-1][2] + prices[i]
        #dp[i][2] = max(dp[i-1][2], dp[i-1][0] - prices[i])
        n = len(prices)

        dp = [[0] * 3 for _ in range(n)]
        dp[0][0] = 0
        dp[0][1] = 0
        dp[0][2] = -prices[0]
        
        for i in range(1, n):
            dp[i][0] = max(dp[i-1][0], dp[i-1][1])
            dp[i][1] = dp[i-1][2] + prices[i]
            dp[i][2] = max(dp[i-1][2], dp[i-1][0] - prices[i])
        return max(dp[n-1][0], dp[n-1][1])

lc713 买卖股票的最佳时机+手续费

卖出的时候扣除手续费

class Solution:
    def maxProfit(self, prices: List[int], fee: int) -> int:
        #dp[i][0]: 第i天不持有，最大利润
        #dp[i][1]: 第i天持有，最大利润

        #dp[i][0] = max(dp[i-1][0], dp[i-1][1] + prices[i] - fee)
        #dp[i][1] = max(dp[i-1][1], dp[i-1][0] - prices[i])

        n = len(prices)

        dp = [[0] * 2 for _ in range(n)]
        dp[0][0] = 0
        dp[0][1] = -prices[0]

        for i in range(1, n):
            dp[i][0] = max(dp[i-1][0], dp[i-1][1] + prices[i] - fee)
            dp[i][1] = max(dp[i-1][1], dp[i-1][0] - prices[i])
        return dp[n-1][0]