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

标签（空格分隔）： leetcode array medium

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题目

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 = 2Output: 8Explanation: 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.

思路

本题是个典型的股票题，由于股票可以买卖多次，因此我们定义两个变量：s0,s1;其中s0和s1的定义如下：

• s0：代表买完股票的利润
• s1：代表卖完股票的利润

更新规则：
update s0 by selling the stock from s1, so s0 = max(s0, s1+p);
update s1 by buying the stock from s0, so s1 = max(s1, s0-p-fee);

代码

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