188. Best Time to Buy and Sell Stock IV Leetcode Python

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本文介绍了一种基于动态规划的算法，用于解决股票交易问题，即在最多进行k次交易的情况下，如何最大化利润。算法通过维护两个向量来跟踪每日最大利润和在特定卖出日的最大利润，最终得出最大可能收益。

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most k transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

Based on Dynamic planning

maintain two vectors: gpro: to day i the maximum profit

lpro: to day i the maximum profit with jth sell by day i

the complexity of this problem is O(k*n)

code is as follow:

class Solution:
    # @return an integer as the maximum profit 
    def helper(self, prices):
        pro = 0
        for i in range(len(prices) - 1):
            pro = max(pro, pro + prices[i+1] - prices[i])
        return pro
    def maxProfit(self, k, prices):
        m = len(prices)
        if m == 0:
            return 0
        if k >= m:### if k >=m this problem become best time to sell II
            return self.helper(prices)
        lpro = [0] * (k + 1)
        gpro = [0] * (k + 1)
        for i in range(len(prices) - 1):
            dif = prices[i + 1] - prices[i]
            j = k
            while j >= 1:
                lpro[j] = max(gpro[j-1]+max(0,dif), lpro[j] + dif)
                gpro[j] = max(gpro[j], lpro[j])
                j-=1
        return gpro[k]