714. Best Time to Buy and Sell Stock with Transaction Fee

#### QUESTION:

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 = 1Selling at prices = 8Buying at prices = 4Selling at prices = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
``````

Note:

`0 < prices.length <= 50000`.

`0 < prices[i] < 50000`.

`0 <= fee < 50000`.

#### EXPLANATION:

1.我们手上持有股票 A[I]

2.我们手上不持有股票 B[I]

#####A[I]：

1.我们在第i天买入的，那么A[i] = B[i-1]-price[i]

2.我们在第i天持有，那么A[i] = A[i-1]

##### B[I]

1.第i天将手上的股票卖出去了 B[i] = A[i-1]+prict[i]-fee

2.继续保持没有股票 B[i] = B[i-1]

#### SOLUTION:

``````class Solution {
public int maxProfit(int[] prices, int fee) {
int cash = 0, hold = -prices;
for (int i = 1; i < prices.length; i++) {
cash = Math.max(cash, hold + prices[i] - fee);
hold = Math.max(hold, cash - prices[i]);
}
return cash;
}
}
``````
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