Leetcode 309. Best Time to Buy and Sell Stock with Cooldown-优快云博客

本文链接：https://blog.youkuaiyun.com/u010370157/article/details/74276136

本文介绍了一种股票交易算法，旨在通过多次买卖操作实现最大利润。该算法遵循特定规则，如不允许同时进行多笔交易且卖出后需冷却一天才能再次购买。通过状态转移方程实现了最优解。

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问题描述

Say you have an array for which the i^th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:

You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)

Example:

prices = [1, 2, 3, 0, 2]
maxProfit = 3
transactions = [buy, sell, cooldown, buy, sell]

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

问题分析：
题意：每次只能买入一个股票，每个时间点最多只能持有一个股票，当卖出股票之后必须歇一天之后重新进行交易。
如果问共有多少种交易，则是卡特兰数。然而问题问的是能够获得的最大收益。
如果前一天持有股票，则今天可以卖掉，如果前一天不持有股票，则判断前一天是否是cooldown。所以必须知道两个信息，即什么时候购买股票，什么时候卖掉股票。
用两个数组：sells[i]表示第i天持有股票，buys[i]表示第i天不持有股票。
状态转移方程：
sells[i]=Max(sells[i-1],buys[i-1]+prices[i])
buys[i]=Max(buys[i-1],sells[i-2]-prices[i])

代码如下：

    public int maxProfit(int[] prices) {
        int n=prices.length;
        if(n<=1)
            return 0;
        int[]sells=new int[n];
        int[]buys=new int[n];
        sells[1]=Math.max(0,prices[1]-prices[0]);
        buys[0]=-prices[0];
        buys[1]=Math.max(-prices[0],-prices[1]);
        for(int i=2;i<n;i++){
            sells[i]=Math.max(buys[i-1]+prices[i],sells[i-1]);
            buys[i]=Math.max(sells[i-2]-prices[i],buys[i-1]);
        }
        return sells[n-1];
    }