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] = 1Selling at prices[3] = 8Buying at prices[4] = 4Selling at prices[5] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
Note:
0 < prices.length <= 50000.
0 < prices[i] < 50000.
0 <= fee < 50000.
本来以为是dp单调队列优化，被题意误导了，小哥哥说是自己阿里面试题，都把dp公式给我了，自己初始状态想了半天。
太太太菜了QAQ
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int sold[50005],hold[50005];
memset(sold,0,sizeof(sold));
memset(hold,0,sizeof(hold));
sold[0]=0;
hold[0]=-prices[0];
int n=prices.size();
for(int i=1;i<n;i++)
{
sold[i]=max(sold[i-1],hold[i-1]+prices[i]-fee);
hold[i]=max(hold[i-1],sold[i-1]-prices[i]);
}
return sold[n-1];
}
};