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 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, prices):
"""
:type k: int
:type prices: List[int]
:rtype: int
"""
if k<1 or len(prices)<2:
return 0
if k>len(prices):
res = 0
for i in range(1,len(prices)):
if prices[i]>prices[i-1]:
res += prices[i] - prices[i-1]
return res
buy = [-100000 for i in range(k)]
sell = [0 for i in range(k)]
for i in prices:
buy[0] = max(buy[0],-i)
sell[0] = max(sell[0],buy[0]+i)
for j in range(1,k):
buy[j] = max(buy[j],sell[j-1]-i)
sell[j] = max(sell[j],buy[j]+i)
return sell[-1]
一开始提交TLE，发现给了个k=1000000000，len(prices)只有10000，这种情况下超时了。这种情况下就等于不限制交易次数，可以简化之。
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