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Leetcode 123. Best Time to Buy and Sell Stock III

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 two transactions.

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

Example 1:

Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Answer: (weijiac ’s answer)

到當前價格,有四種情況:

股票2:

     a.在當前價格買入或不買入,買入時手中的錢已經有股票1所賺的錢,hold2=max(release1-p,hold2)

     b.在當前價格賣出或不賣出,release2=max(hold2+p,release2)

股票1:

     a.在當前價格買入或不買入,hold1=max(-p,hold1)

     b.在當前價格賣出或不賣出,release1=max(hold1+p,release1)

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        hold1=-99999999
        hold2=-99999999
        release1=0
        release2=0
        
        for p in prices:
            release2=max(hold2+p,release2)
            hold2=max(release1-p,hold2)
            release1=max(hold1+p,release1)
            hold1=max(-p,hold1)
        return release2