topic
Given an integer array prices, where prices[i] represents the stock price on the ith day; the integer fee represents the transaction fee for the stock.
You can complete an unlimited number of transactions, but you will need to pay a fee for each transaction. If you've already bought a stock, you can't keep buying it until you sell it.
Returns the maximum profit earned.
Note: A transaction here refers to the entire process of buying, holding and selling stocks, and you only need to pay a handling fee for each transaction.
Example 1:
Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
Output: 8
Explanation: Maximum profit that can be achieved:
buy here prices[0] = 1
sell here prices[ 3] = 8
buy here prices[4] = 4
sell here prices[5] = 9
Total profit: ((8 - 1) - 2) + ((9 - 4) - 2) = 8
Example 2:
Input: prices = [1,3,7,5,10,3], fee = 3
Output: 6
Source: LeetCode
Link: https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee
answer
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
//动态规划
vector<vector<int> > dp(prices.size(), vector<int>(2,0));
dp[0][0] = 0;
dp[0][1] = -prices[0]-fee;
for(int i = 1; i < prices.size(); i++)
{
dp[i][0] = max(dp[i-1][0],dp[i-1][1] + prices[i]);
dp[i][1] = max(-prices[i]- fee + dp[i-1][0], dp[i-1][1]);
}
return dp[prices.size()-1][0];
}
};
result
