码迷,mamicode.com
首页 > 其他好文 > 详细

动态规划_leetcode309(状态好难定义,又是买彩票)

时间:2019-03-17 13:26:23      阅读:239      评论:0      收藏:0      [点我收藏+]

标签:obj   动态规划   def   ash   fit   span   递推公式   状态   style   

class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""


length = len(prices)

if length <= 1:
return 0

hold = [-1 for i in range(length)]
cash = [0 for i in range(length)]

hold[0] = -prices[0]
hold[1] = max(hold[0],-prices[1])
cash[1] = max(cash[0],hold[0] + prices[1])

for i in range(2,length):
cash[i] = max(cash[i-1],hold[i-1]+prices[i])
hold[i] = max(hold[i-1],cash[i-2]-prices[i])



return cash.pop()




# 好难
# sell[i]表示截至第i天,最后一个操作是卖时的最大收益;
# buy[i]表示截至第i天,最后一个操作是买时的最大收益;
# cool[i]表示截至第i天,最后一个操作是冷冻期时的最大收益;
# 递推公式:
# sell[i] = max(buy[i-1]+prices[i], sell[i-1]) (第一项表示第i天卖出,第二项表示第i天冷冻)
# buy[i] = max(cool[i-1]-prices[i], buy[i-1]) (第一项表示第i天买进,第二项表示第i天冷冻)
# cool[i] = max(sell[i-1], buy[i-1], cool[i-1])
# python:

class Solution2:
def maxProfit(self, prices):
n = len(prices)
if n == 0:
return 0
sell = [0 for _ in range(n)]
buy = [0 for _ in range(n)]
cool = [0 for _ in range(n)]
buy[0] = -prices[0]
for i in range(1,n):
sell[i] = max(buy[i-1] + prices[i], sell[i-1])
buy[i] = max(cool[i-1] - prices[i], buy[i-1])
cool[i] = max(sell[i-1], buy[i-1],cool[i-1])
return sell[-1]

动态规划_leetcode309(状态好难定义,又是买彩票)

标签:obj   动态规划   def   ash   fit   span   递推公式   状态   style   

原文地址:https://www.cnblogs.com/lux-ace/p/10546577.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!