123. Best Time to Buy and Sell Stock III 买卖股票的最佳时机 III

作者： 负雪明烛 id： fuxuemingzhu 个人博客： http://fuxuemingzhu.cn/

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题目地址：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/

题目描述

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.

题目大意

给出了一堆股票价格，最多做两次交易，求最大的收益。

解题方法

这个题太难了，看了一个多小时没看懂。直接抄的Gradyang的做法，罪过罪过。

地址：http://www.cnblogs.com/grandyang/p/4281975.html

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices: return 0
        N = len(prices)
        g = [[0] * 3 for _ in range(N)]
        l = [[0] * 3 for _ in range(N)]
        for i in range(1, N):
            diff = prices[i] - prices[i - 1]
            for j in range(1, 3):
                l[i][j] = max(g[i - 1][j - 1] + max(diff, 0), l[i - 1][j] + diff)
                g[i][j] = max(l[i][j], g[i - 1][j])
        return g[-1][-1]

日期

2018 年 11 月 29 日 —— 时不我待