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

题目原文

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.

题意分析

给定一组某一stock在每一天的价格，买卖次数不限，每次买入必须在卖出之后，且每次卖出时都需要fee的手续费，求解最大的收益。

解法分析

本题采用两种方法解

• 动态规划
• 贪心算法

动态规划

对于第i天的最大收益，应分成两种情况，一是该天结束后手里没有stock，可能是保持前一天的状态也可能是今天卖出了，此时令收益为cash；二是该天结束后手中有一个stock，可能是保持前一天的状态，也可能是今天买入了，用hold表示。由于第i天的情况只和i-1天有关，所以用两个变量cash和hold就可以，不需要用数组。C++代码如下：

class Solution { public: int maxProfit(vector<int>& prices, int fee) { int cash=0;//the maxPro you have if you don't have a stock that day int hold=-prices[0];//the maxPro you have if you have a stock that day, if you have a stock the first day,hold=-prices[0] int i; for(i=1;i<prices.size();i++){ cash=max(cash,hold+prices[i]-fee);//cash in day i is the maxvalue of cash in day i-1 or you sell your stock hold=max(hold,cash-prices[i]); } return cash; } };

这里需要注意cash和hold的初始值，最终输出cash，因为最后一天的情况一定是手里没有stock的。

贪心算法

贪心选择的关键是找到一个最大后是不是能够卖掉stock，重新开始寻找买入机会。比如序列1 3 2 8，如果发现2小于3就完成交易买1卖3，此时由于fee=2，（3-1-fee）+(8-2-fee)<(8-1-fee)，所以说明卖早了，令max是当前最大price，当（max-price[i]>=fee）时可以在max处卖出，且不会存在卖早的情况，再从i开始重新寻找买入机会。c++代码如下：

class Solution { public: int maxProfit(vector<int>& prices, int fee) { int profit=0; int curProfit=0; int minP=prices[0]; int maxP=prices[0]; int i; for(i=1;i<prices.size();i++){ minP=min(minP,prices[i]); maxP=max(maxP,prices[i]); curProfit=max(curProfit,prices[i]-minP-fee); if((maxP-prices[i])>=fee){//can just sell the stock at maxP day. profit+=curProfit; curProfit=0; minP=prices[i]; maxP=prices[i]; } } return profit+curProfit;//the last trade have to be made if there is some profit } };

curProfit记录了当前一次交易能得到的最大收益，只有当maxP-prices[i]>=fee时，才将curProfit累加到总的收益中。最后一次交易不需要考虑是否早卖了，所以直接累加最后一次的curProfit。