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

714. 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] = 1

Selling at prices[3] = 8

Buying at prices[4] = 4

Selling at prices[5] = 9

The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Note:

-

0 < prices.length <= 50000

.

-

0 < prices[i] < 50000

.

-

0 <= fee < 50000

.

题目大意

题中给出物品不同时间的不同售价

prices

，购买时需要支付

fee

，要求计算买入卖出后，你能获得的最大利润。

解题思路

每一天一共可以有3个操作，分别是买、卖、不买不卖。

策略:

1. 买入的最大累积收益，决定买或不买。

2. 卖出的最大累积收益，决定卖或不卖。

算法复杂度

O(n)

代码实现

class Solution {
public:
int max(int a, int b) {
return (a > b) ? a : b;
}
int maxProfit(vector<int>& prices, int fee) {
const int minVal = -100001;
int days = prices.size();
vector<int> possess(days+1, 0);
vector<int> profit(days+1, 0);
possess[0] = minVal;
for (int i = 1; i < days+1; ++i) {
// not buy or buy
possess[i] = max(possess[i-1], profit[i-1]-prices[i-1]-fee);
// not sell or sell
profit[i] = max(profit[i-1], possess[i-1]+prices[i-1]);
}
return profit[days];
}
};