309. Best Time to Buy and Sell Stock with Cooldown

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 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]
思路：主要本题是有冰冻期的，是一个dp的题，设置两个数组buy[i]表示在第i天买了，sell[i]表示在第i天卖了之后的最大利润。
dp[i]=max(dp[i-1],sell[i-2]-prices[i]) sell[i]=max(sell[i-1],buy[i-1]+prices[i]);
根据此递推式可以得到AC代码
代码如下(已通过leetcode)
public class Solution {    public int maxProfit(int[] prices) {        int n=prices.length;        if(n==0) return 0;        int[] sell=new int[n];        int[] buy=new int[n];        sell[0]=0;        buy[0]=-prices[0];        for(int i=1;i<n;i++) {        sell[i]=Math.max(sell[i-1], buy[i-1]+prices[i]);        if(i>=2) buy[i]=Math.max(buy[i-1], sell[i-2]-prices[i]);        else buy[i]=Math.max(buy[i-1], -prices[i]);        }        return sell[n-1];    }}