LeetCode 714. Best Time to Buy and Sell Stock with Transaction Fee

原题链接在这里：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/

题目：

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.

Note:

0 < prices.length <= 50000.
0 < prices[i] < 50000.
0 <= fee < 50000.

题解：

Let buy[i] denotes maximum profit till index i, ending with buy.

Let sell[i] denotes maximum profit till index i, ending with sell.

buy[i] = max(buy[i-1], sell[i-1]-prices[i]).

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

Use variable instead of array to save space.

Note: Do NOT forget to update variable after each iteration.

Time Complexity: O(n). n = prices.length.

Space: O(1).

 1 class Solution {
 2     public int maxProfit(int[] prices, int fee) {
 3         if(prices == null || prices.length < 2){
 4             return 0;
 5         }
 6         
 7         int b0 = -prices[0];
 8         int b1 = b0;
 9         int s0 = 0;
10         int s1 = 0;
11         for(int i = 1; i<prices.length; i++){
12             b0 = Math.max(b1, s1-prices[i]);
13             s0 = Math.max(s1, b1+prices[i]-fee);
14             
15             b1 = b0;
16             s1 = s0;
17         }
18         
19         return s0;
20     }
21 }

T[i][k][0] 代表maximum profit that could be gained at the end of the i-th day with at most k transactions. 最后手上剩下0股stock.

递推公式就是

T[i][k][0] = max(T[i-1][k][0], T[i-1][k][1] + prices[i])
T[i][k][1] = max(T[i-1][k][1], T[i-1][k-1][0] - prices[i])

如果需要加上transaction fee就变成

T[i][k][0] = max(T[i-1][k][0], T[i-1][k][1] + prices[i])
T[i][k][1] = max(T[i-1][k][1], T[i-1][k][0] - prices[i] - fee)

买入时交fee.

最后肯定是卖掉手上的股票收益更多，所以返回T[i][k][0].

Time Complexity: O(n). n = prices.length.

Space: O(1).

AC Java:

 1 class Solution {
 2     public int maxProfit(int[] prices, int fee) {
 3         int tIk0 = 0;
 4         int tIk1 = Integer.MIN_VALUE;
 5         for(int price : prices){
 6             int preTransactionIk0 = tIk0;
 7             tIk0 = Math.max(tIk0, tIk1+price);
 8             tIk1 = Math.max(tIk1, tIk0-price-fee);
 9         }
10         return tIk0;
11     }
12 }

买卖股票类题目都可以套用这个思路.

Reference: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/discuss/108870/