Description
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] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
Note:
-
0 < prices.length <= 50000.*0 < prices[i] < 50000.*0 <= fee < 50000.
Solution
DP, time O(n), space O(n)
需要用两个数组来DP，只使用一个数组的话时间复杂度为O(n^2)会TLE。根据在第i天持有或不持有stock来分开DP。
class Solution {
public int maxProfit(int[] prices, int fee) {
int n = prices.length;
if (n < 2) {
return 0;
}
//Till day i, the max profit is hold[i] if I hold the stock.
int[] hold = new int[n];
//Till day i, the max profit is notHold[i] if I do not hold the stock.
int[] notHold = new int[n];
hold[0] = -prices[0];
for (int i = 1; i < n; ++i) {
hold[i] = Math.max(hold[i - 1], notHold[i - 1] - prices[i]);
notHold[i] = Math.max(notHold[i - 1]
, hold[i - 1] + prices[i] - fee);
}
return notHold[n - 1];
}
}
