leetcode011-Best Time to Buy and Sell Stock with Transaction Fee

Your are given an array of integers 

prices

,
for which the 

i

-th element is the price of a given stock on day 

i

;
and a non-negative integer 

fee

 representing a transaction fee.
You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you
buy again.)
Return the maximum profit you can make.
Example 1:

Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
Output: 8
Explanation: The maximum profit can be achieved by:

Buying at prices[0] = 1Selling at prices[3] = 8Buying at prices[4] = 4Selling at prices[5] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Note:

0 < prices.length <= 50000

.

0 < prices[i] < 50000

.

0 <= fee < 50000

.

解题思路：使用动态规划，每天能有三个可能的操作，手上有股票能卖出，手上没有股票可以买入，可以什么都不做。
1.用hold[i]表示第i天，我要持有股票时的最大收益。那么我可能从之前的第k(k<i)天时就开始持有股票到i-1天时依然持有，也可能是第i-1天时我手上没有股票于是今天买入。（注意，hold[i]表示第i天完成交易事务后，我手上持有股票）择两种可能中的最大值，为第i天持有股票的最大收益。
2.用unhold[i]表示第i天，我手上没有股票时的最大收益。那么我可能从之前的第k(k<i)天时就没有股票到i-1天时依然没有，也可能是第i-1天时我手上持有股票于是今天卖出。（注意，unhold[i]表示第i天完成交易事务后，我手上没有股票）择两种可能中的最大值，为第i天没有股票时的最大收益。
3.转移方程：hold[i] = max{hold[i-1](第i-1天结束我持股),unhold[i-1](第i-1天结束我不持股)-prices[i]}
     unhold[i] = max{unhold[i-1],hold[i-1]-fee+prices[i]}
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int days = prices.size();
vector<int> hold(days);
vector<int> unhold(days);
hold[0] = -prices[0];
unhold[0] = 0;
for (int i = 1; i < days; i++) {
hold[i] = max(hold[i-1],unhold[i-1]-prices[i]);
unhold[i] = max(unhold[i-1],hold[i-1]-fee+prices[i]);
}
return unhold[days-1];
}
};