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

题目描述

Best Time to Buy and Sell Stock with Transaction Fee

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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.

分析

这也是一道DP问题。

在第i天,只会有两个状态,手上现在有股票,和手上现在没有股票。分别计算着两者当前获利的最大值。

没有股票 = max(没有股票,卖出股票)

持有股票 = max(持有股票,买进股票)

注意这里买进股票是需要手续费的

对应的原来的Best Time to Buy and Sell Stock II也可以用这种方法求解,只是不用减去手续费了。

AC代码

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class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int s0 = 0, s1 = INT_MIN;
for(int p:prices) {
int tmp = s0;
s0 = max(s0, s1+p);
s1 = max(s1, tmp-p-fee);
}
return s0;
}

};