309.Best Time to Buy and Sell Stock with Cooldown
===

###### tags: `Medium`,`Array`,`DP`

[309. Best Time to Buy and Sell Stock with Cooldown](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/)

### 題目描述

You are given an array `prices` where `prices[i]` is the price of a given stock on the i^th^ 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).

### 範例

**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` <= 5000
* 0 <= `prices[i]` <= 1000

### 解答

#### C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int buy = INT_MIN, hold = INT_MIN, sell = 0, cooldown = 0;
        for (auto price : prices) {
            hold = max(hold, buy);
            buy = cooldown - price;
            cooldown = max(sell, cooldown);
            sell = hold + price;
        }
        return max(cooldown, sell);
    }
};

> [name=Yen-Chi Chen][time=Fri, Dec 23, 2022]

#### Python

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        buy, hold, sell, cooldown = -math.inf, -math.inf, 0, 0
        for price in prices:
            hold = max(hold, buy)
            buy = cooldown - price
            cooldown = max(cooldown, sell)
            sell = hold + price
        return max(sell, cooldown)

> [name=Yen-Chi Chen][time=Fri, Dec 23, 2022]

### 三個狀態的做法


class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        hold, sold, rest = -math.inf, 0, 0
        for price in prices:
            hold, sold, rest = max(hold, rest - price), hold + price, max(rest, sold)
        return max(sold, rest)

> [name=Yen-Chi Chen][time=Mon, Dec 26]

### Reference

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