- LeetCode 121. Best Time to Buy and Sell Stock–Java，Python，C++解法
- LeetCode 122. Best Time to Buy and Sell Stock II–贪心–Java,C++,Python解法
- LeetCode 123. Best Time to Buy and Sell Stock III–Python解法–动态规划–数学题
- LeetCode 309. Best Time to Buy and Sell Stock with Cooldown–Java解法-卖股票系列题目
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.
这道题目看起来跟这道很像：LeetCode 122. Best Time to Buy and Sell Stock II–贪心–Java,C++,Python解法，但其实不一样。
class Solution: def maxProfit(self, prices: List[int]) -> int: if prices == : return 0 buyOne = 0x7ffffff SellOne = 0 buyTwo = 0x7ffffff SellTwo = 0 for p in prices: buyOne = min(buyOne, p) SellOne = max(SellOne, p - buyOne) buyTwo = min(buyTwo, p - SellOne) SellTwo = max(SellTwo, p - buyTwo) return SellTwo
[1, 3, 2, 7]为例：