算法: 买卖股票的最佳时机 III123. Best Time to Buy and Sell Stock III

You are given an array prices where prices[i] is the price of a given stock on the ith day.

Find the maximum profit you can achieve. You may complete at most two transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

Input: prices = [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: prices = [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: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Constraints:

• 1 <= prices.length <= 105
• 0 <= prices[i] <= 105

两头逼近算法

用下面的例子解析

2 8 4 5 6 
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遍历两次

1. 从左到右，先用数组profits记录当前位置最大的利润。这里的技巧是要记录最小值minPrice。
2. 从右到左，计算最大值利润maxPrice - prices[i]，技巧是记录最大值maxPrice。然后记录两个方向的最大值max(res, maxProfit + profits[i])

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        n = len(prices)
        if n < 2:
            return 0
        profits = []
        minPrice = prices[0]
        maxProfit = 0
        for price in prices:
            minPrice = min(minPrice, price)
            maxProfit = max(maxProfit, price - minPrice)
            profits.append(maxProfit)
        
        maxProfit = 0
        maxPrice = prices[-1]
        res = 0
        for i in range(n-1, -1, -1):
            maxPrice = max(maxPrice, prices[i])
            maxProfit = max(maxProfit, maxPrice - prices[i])
            res = max(res, maxProfit + profits[i])
            
        return res