Leetcode算法学习日志-714 Best Time to Buy and Sell Stock with Transaction Fee

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 thei-th element is the price of a given stock on day i; and a non-negative integerfee 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] = 9The 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。