weekly contest 55 Best Time to Buy and Sell Stock with Transaction Fee

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] = 1Selling at prices[3] = 8Buying at prices[4] = 4Selling at prices[5] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

---

分析

---

这题和之前的冷静的买卖袜子是一个题目

那我们先定义状态

buy[i] 表示第i天的买收益的最大

sell[i] 表示第i天的卖收益的最大

接着填写初时状态

buy[0] = - prices[0] ( 我们买了第一件）

sell[0] = 0 （没法卖，没有收益）

接着是状态转移

buy[i] = max( sell[i-1] - prices[i], buy[i-1]);
( 前一天卖了后今天买了， 前一天的最大）

sell[i] = max( buy[i-1] + prices[i] - fee , sell[i-1]);

---

代码

---

class Solution {public:    int maxProfit(vector<int>& prices, int fee) {        // buy[i] = max(sell[i-1]-price, buy[i-1])        // sell[i] = max(buy[i-1]+price, sell[i-1])        int n = prices.size() ;         if( n <= 1 )            return 0 ;        vector<int> buy ;        vector<int> sell;        buy.resize(n , 0);        sell.resize( n , 0);        buy[0] = - prices[0];        sell[0] = 0 ;        for( int i= 1 ; i<prices.size() ; i++){            buy[i] = max( sell[i-1] - prices[i] , buy[i-1] ) ;            sell[i] = max( buy[i-1] + prices[i] - fee , sell[i-1] ) ;        }        return sell[n-1];    }};

---

时间复杂度

O(N)

空间复杂度

O(N)