|
马上注册,结交更多好友,享用更多功能^_^
您需要 登录 才可以下载或查看,没有账号?立即注册
x
本帖最后由 Seawolf 于 2020-9-10 22:33 编辑
Say you have an array for which the i-th element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most k transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
Example 1:
Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
Example 2:
Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
- class Solution:
- def maxProfit(self, k: int, prices: List[int]) -> int:
- #hold, sold
- if k <= 0 or prices == None or len(prices) == 0:
- return 0
-
- #the maximum trahsanction time is less than k
- if 2 * k > len(prices):
- res = 0
- for i in range(1, len(prices)):
- res += max(0, prices[i] - prices[i - 1])
- return res
-
- #regular case
- dp = [[-math.inf, 0] for _ in range(k)]
-
- for price in prices:
- temp = dp
- for i in range(k):
- if i == 0:
- dp[i][0] = max(temp[i][0], 0 - price)
- dp[i][1] = max(temp[i][1], temp[i][0] + price)
- else:
- dp[i][0] = max(temp[i][0], temp[i - 1][1] - price)
- dp[i][1] = max(temp[i][1], temp[i][0] + price)
-
- return dp[-1][-1]
复制代码 |
|