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/
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).
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.
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.
Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.
class Solution(object): def maxProfit(self, prices): """ :type prices: List[int] :rtype: int """ if not prices: return 0 N = len(prices) g = [ * 3 for _ in range(N)] l = [ * 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]
2018 年 11 月 29 日 —— 时不我待