[Array]123. Best Time to Buy and Sell Stock III

• 分类：Array/Greedy
• 时间复杂度: O(n)
• 空间复杂度: O(n)

123. Best Time to Buy and Sell Stock III

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.

代码：

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if prices==None or len(prices)<2:
            return 0
        
        profits=[[0 for i in range(len(prices))] for i in range(2)]   
        
        buy_price=prices[0]
        for i in range(1,len(prices)):
            buy_price=min(prices[i],buy_price)
            profits[0][i]=max(profits[0][i-1],prices[i]-buy_price)
            
        sell_price=prices[-1]
        for i in range(len(prices)-2,-1,-1):
            sell_price=max(prices[i],sell_price)
            profits[1][i]=max(profits[1][i+1],sell_price-prices[i])
            
        total_profit=0
        for i in range(len(profits[0])-1):
            total_profit=max(total_profit,profits[0][i]+profits[1][i+1])
        total_profit=max(total_profit,profits[0][-1])
        
        return total_profit

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讨论：

1.为啥别人写的代码如此优秀，思路清晰。。。我感觉我好傻啊