309.leetcode Best Time to Buy and Sell Stock with Cooldown(medium)[动态规划]

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]

需要一天的冷冬天，这个题采用两个状态数组来完成，分别是holdDp:表示在第i天没有持股的情况下获得的最大受益，

此时有两种情况，一种是当天是冷冻期所有holdDp为i-1天的受益，还有一种是今天的股票卖了，收益是prices[i]+nodp[i-1]，取两者中的最大值；

nodp表示当天持股的时候获得的最大受益，分别是今天与昨天一样有股票没卖nodp[i-1],或者是今天买了股票，那么前天才买了股票，因此是holdDp[i-2]-prices[i]

注意初始化的时候nodp[0] = -prices[0]

class Solution {
public:
int getMax(int a,int b)
{
if(a>=b) return a;
else return b;
}
int maxProfit(vector<int>& prices) {
//这个题采用两个状态数组来完成，分别是holdDp:表示在第i天没有持股的情况下获得的最大受益，此时有两种情况，一种是当天是冷冻期所有holdDp为i-1天的受益，还有一种是今天的股票卖了，收益是prices[i]+nodp[i-1]，取两者中的最大值；nodp表示当天持股的时候获得的最大受益，分别是今天与昨天一样有股票没卖nodp[i-1],或者是今天买了股票，那么前天才买了股票，因此是holdDp[i-2]-prices[i]int n = prices.size();
if(n<=1) return 0;
int holdDp
;
int noDp
;
holdDp[0] = 0;
noDp[0] = -prices[0];
for(int i=1;i<n;i++)
{
holdDp[i] = getMax(holdDp[i-1],noDp[i-1]+prices[i]);
if(i>2)
noDp[i] = getMax(noDp[i-1],holdDp[i-2]-prices[i]);
else
noDp[i] = getMax(noDp[i-1],-prices[i]);
}
return holdDp[n-1];
}
};