Problem Description

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]

Some Details

• 標準的簡單DP題目

Solution

DP沒怎麼切，這題雖然難度不高但是既然切到了就趕快寫了，畢竟作業難以完成= =。
買入，賣出，冷卻，三狀態，ok
f[i][0] 第i天手上沒貨的最優獲利
f[i][1] 第i天手上有貨的最優獲利
f[i][2] 第i天進入冷卻的最優獲利
轉移方程：
f[i][0]=max(f[i-1][0],f[i-1][2])
//前一天就沒貨或者前一天冷卻
f[i][1]=max(f[i-1][0]-price[i],f[i-1][1])
//前一天就有貨或者今天買入
f[i][2]=f[i-1][1]+price[i];
//前一天有貨今天賣了
結束思密達。

Code

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int f[10100][4],siz=prices.size();
        if (siz==0) return 0;
        f[0][0]=0;
        f[0][1]=prices[0]*-1;
        f[0][2]=0;
        for (int i=1;i<siz;i++)
        {
            f[i][0]=max(f[i-1][0],f[i-1][2]);
            f[i][1]=max(f[i-1][0]-prices[i],f[i-1][1]);
            f[i][2]=f[i-1][1]+prices[i];
        }
        int ans=f[siz-1][0];
        ans=max(ans,f[siz-1][1]);
        ans=max(ans,f[siz-1][2]);
        return ans;
    }
};