- 分類：Array/Greedy
- 時間復雜度: O(n)
- 空間復雜度: O(n)
123. Best Time to Buy and Sell Stock III
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.
代碼：
class Solution:
def maxProfit(self, prices: List[int]) -> int:
if prices==None or len(prices)<2:
return 0
profits=[[0 for i in range(len(prices))] for i in range(2)]
buy_price=prices[0]
for i in range(1,len(prices)):
buy_price=min(prices[i],buy_price)
profits[0][i]=max(profits[0][i-1],prices[i]-buy_price)
sell_price=prices[-1]
for i in range(len(prices)-2,-1,-1):
sell_price=max(prices[i],sell_price)
profits[1][i]=max(profits[1][i+1],sell_price-prices[i])
total_profit=0
for i in range(len(profits[0])-1):
total_profit=max(total_profit,profits[0][i]+profits[1][i+1])
total_profit=max(total_profit,profits[0][-1])
return total_profit
討論：
1.為啥別人寫的代碼如此優(yōu)秀，思路清晰浩嫌。退唠。。我感覺我好傻啊
