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

题目

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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.

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分析

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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]);

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代码

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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 ; i1] - prices[i] , buy[i-1] ) ;
            sell[i] = max( buy[i-1] + prices[i] - fee , sell[i-1] ) ;
        }
        return sell[n-1];
    }
};

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时间复杂度

O(N)

空间复杂度

O(N)