算法: 冷却时间买卖股票的最佳时机309. Best Time to Buy and Sell Stock with Cooldown
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 as many transactions as you like (i.e., buy one and sell one share of the stock multiple times) with the following restrictions:
After you sell your stock, you cannot buy stock on the next day (i.e., cooldown one day).
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
Example 1:
Input: prices = [1,2,3,0,2]
Output: 3
Explanation: transactions = [buy, sell, cooldown, buy, sell]
Example 2:
Input: prices = [1]
Output: 0
Constraints:
1 <= prices.length <= 50000 <= prices[i] <= 1000
状态转移解法
关键是状态转换的 3 个状态和 5 个边。3 个状态是notHold (stock)、hold (stock)和notHold_cooldown。后两者的初始值为负无穷大，因为它们没有意义，即你一开始不会持有股票，一开始也没有冷却时间。5条边：
hold -----do nothing----->hold
hold -----sell----->notHold_cooldown
notHold -----do nothing -----> notHold
notHold -----buy-----> hold
notHold_cooldown -----do nothing----->notHold
class Solution:
def maxProfit(self, prices: List[int]) -> int:
notHold, notHoldCoolDown, hold = 0, float('-inf'), float('-inf')
for p in prices:
hold, notHold, notHoldCoolDown = max(hold, notHold - p), max(notHold, notHoldCoolDown), hold + p
return max(notHold, notHoldCoolDown)
参考
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/discuss/75942/4-line-Python-solution-52-ms
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/discuss/761981/PythonGo-O(n)-by-DP-and-state-machine.-w-Visualization
