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 (ie, you must sell the stock before you buy again). 这道题目可以采用两种方法来解，第一种解法是 分别从左往右扫描一次 以及从右往左扫描一次依次把得到的最大的maxpro 放到数组里面 stack1 stack2 最后求解的时候 将stack1[i]+stack2[i+1] 得到的最大值就是最大值。代码如下

class Solution:
    # @param prices, a list of integer
    # @return an integer
    def maxProfit(self, prices):
        if len(prices)==0:
            return 0
        stack1=[]
        stack2=[]
        lowprice=prices[0]
        maxpro=0
        highprice=prices[-1]
        for index in range(len(prices)):
            lowprice=min(lowprice,prices[index])
            maxpro=max(maxpro,prices[index]-lowprice)
            stack1.append(maxpro)
        maxpro=0
        for index in reversed(range(len(prices))):
            highprice=max(highprice,prices[index])
            maxpro=max(maxpro,highprice-prices[index])
            stack2.insert(0,maxpro)
        
        for index in range(len(prices)-1):
            maxpro=max(maxpro,stack1[index]+stack2[index+1])
        return maxpro
            

i表示第i天 j表示一共可以进行j次交易

全局最优的表达式为g[j]=max(g[j],l[j])

局部最优的表达为当前最优和 前一个全局最优加上当前可以挣的钱的最大 表达式为

l[j]=max(g[j-1]+max(dif,0),l[j]+dif)

具体代码如下：

class Solution:
    # @param prices, a list of integer
    # @return an integer
    def maxProfit(self, prices):
        g=[0,0,0]
        l=[0,0,0]
        for i in range(len(prices)-1):
            dif=prices[i+1]-prices[i]
            for j in reversed(range(1,3)):
                l[j]=max(g[j-1]+max(dif,0),l[j]+dif)
                g[j]=max(l[j],g[j])
        return g[2]

Best Time to Buy and Sell Stock III Leetcode Python