188. Best Time to Buy and Sell Stock IV

188. Best Time to Buy and Sell Stock IV

难度: Hard

刷题内容

原题连接

内容描述

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.
Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

解题方案

参考meng789987

思路 1
- 时间复杂度: O(kN^2)- 空间复杂度: O(kN)

dp[k][i]代表在第i天完成第k次交易的最大profit

因此状态方程为dp[k][i] = max(dp[k][i-1], prices[i]-prices[j]+dp[k-1][j-1]),其中0 < j <= i

意思就是在第i天完成第k次交易的最大profit就是,在第i-1天完成第k次交易的最大profit
在j-1天完成前k-1次交易的最大profit(然后在第j天最后一次买入股票,并在第i天卖出两者中的最大值

另外这里还有一个trick就是当K >= len(prices)//2 的时候,实际上就跟无限次交易一样了

但是这样还是超时了

```
class Solution(object):
def maxProfit(self, k, prices):
"""
:type k: int
:type prices: List[int]
:rtype: int
"""
K = k
if K >= len(prices) // 2:
return sum([max(prices[i+1]-prices[i], 0) for i in range(len(prices)-1)])
if not prices or len(prices) == 0:
return 0
dp = [[0] * len(prices) for i in range(K+1)]
for k in range(1, K+1):

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