Best Time to Buy and Sell Stock III Leetcode Python

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


第二种做法是用动态规划的方法解的
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]