# Best Time to Buy and Sell Stock with Transaction Fee

2018-04-11 02:07:13         来源：mbinary

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.   思路 动态规划，弄清楚，f(n),与f(n 1)的关系，确定转移方程，以及当再次加入一组时是否可以多于两次的fee，代码29，30行就是处理这点

## show me the code

```class Solution(object):
def maxProfit(self, prices, fee):
"""
:type prices: List[int]
:type fee: int
:rtype: int
"""
n = len(prices)
if n<2:return 0
li = []
if prices[1]-prices[0]>fee:
li.append([0,1])
idx = -1
else:
if (prices[0]>prices[1]):idx = 1
else:idx = 0
for i in range(2,n):
if not li:
if  prices[i]-prices[idx]>fee:li.append([idx,i])
elif prices[i]prices[last[1]]:
last[1] = i
else:idx = i
else:
if prices[i]>prices[idx] fee:
if prices[last[1]]-prices[idx]prices[last[1]]: last[1]=i
elif prices[i]

```