題目

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 = 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.
 

分析

這題和之前的冷靜的買賣襪子是一個題目

那我們先定義狀態

buy[i] 表示第i天的買收益的最大

sell[i] 表示第i天的賣收益的最大

接著填寫初時狀態

buy[0] = - prices[0] ( 我們買了第一件）

sell[0] = 0 （沒法賣，沒有收益）

接著是狀態轉移

buy[i] = max( sell[i-1] - prices[i], buy[i-1]);
( 前一天賣了後今天買了， 前一天的最大）

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

程式碼

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];
    }
};

時間複雜度 O(N)

空間複雜度O(N)