weekly contest 55 Best Time to Buy and Sell Stock with Transaction Fee
阿新 • • 發佈:2019-01-22
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];
}
};