[leetcode]Best Time to Buy and Sell Stock III @ Python

原题地址：https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/

题意：

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).

解题思路：只允许做两次交易，这道题就比前两道要难多了。解法很巧妙，有点动态规划的意思：开辟两个数组f1和f2，f1[i]表示在price[i]之前进行一次交易所获得的最大利润，f2[i]表示在price[i]之后进行一次交易所获得的最大利润。则f1[i]+f2[i]的最大值就是所要求的最大值，而f1[i]和f2[i]的计算就需要动态规划了，看代码不难理解。

代码：

class Solution:
    # @param prices, a list of integer
    # @return an integer
    def maxProfit(self, prices):
        length=len(prices)
        if length==0: return 0
        f1=[0 for i in range(length)]
        f2=[0 for i in range(length)]

        minV=prices[0]; f1[0]=0
        for i in range(1,length):
            minV=min(minV, prices[i])
            f1[i]=max(f1[i-1],prices[i]-minV)

        maxV=prices[length-1]; f2[length-1]=0
        for i in range(length-2,-1,-1):
            maxV=max(maxV,prices[i])
            f2[i]=max(f2[i+1],maxV-prices[i])

        res=0
        for i in range(length):
            if f1[i]+f2[i]>res: res=f1[i]+f2[i]
        return res