[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： 交易市场的“低买高卖" 法则(buy low and sell high‘ )

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p1和p2，p1[i]表示在price[i]之前进行一次交易所获得的最大利润，

p2[i]表示在price[i]之后进行一次交易所获得的最大利润。

则p1[i]+p2[i]的最大值就是所要求的最大值，

而p1[i]和p2[i]的计算就需要动态规划了，看代码不难理解。

class Solution:
    # @param prices, a list of integer
    # @return an integer
    def maxProfit(self, prices):
        n = len(prices)
        if n <= 1: return 0
        p1 = [0] * n
        p2 = [0] * n

        minV = prices[0]
        for i in range(1,n):
            minV = min(minV, prices[i])       # Find low and buy low
            p1[i] = max(p1[i - 1], prices[i] - minV)

        maxV = prices[-1]
        for i in range(n-2, -1, -1):
            maxV = max(maxV, prices[i])     # Find high and sell high
            p2[i] = max(p2[i + 1], maxV - prices[i])

        res = 0
        for i in range(n):
            res = max(res, p1[i] + p2[i])
        return res