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题目:
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 (ie, you must sell the stock before you buy again).
代码:Runtime: 175 ms
1 class Solution: 2 # @param prices, a list of integer 3 # @return an integer 4 def maxProfit_with_k_transactions(self, prices, k): 5 days = len(prices) 6 local_max = [[0 for i in range(k+1)] for i in range(days)] 7 global_max = [[0 for i in range(k+1)] for i in range(days)] 8 for i in range(1,days): 9 diff = prices[i] - prices[i-1] 10 for j in range(1,k+1): 11 local_max[i][j] = max(local_max[i-1][j]+diff, global_max[i-1][j-1]+max(diff,0)) 12 global_max[i][j] = max(local_max[i][j], global_max[i-1][j]) 13 return global_max[days-1][k] 14 15 def maxProfit(self, prices): 16 if prices is None or len(prices)<2: 17 return 0 18 return self.maxProfit_with_k_transactions(prices, 2)
思路:
不是自己想的,参考这篇博客http://blog.csdn.net/fightforyourdream/article/details/14503469
跟上面博客一样的思路就不重复了,下面是自己的心得体会:
1. 这类题目,终极思路一定是往动态规划上靠,我自己概括为“全局最优 = 当前元素之前的所有元素里面的最优 or 包含当前元素的最优”
2. 这道题的动归的难点在于,只靠一个迭代公式无法完成寻优。
思路如下:
global_max[i][j] = max( global_max[i-1][j], local_max[i][j])
上述的迭代公式思路很清楚:“到第i个元素的全局最优 = 不包含第i个元素的全局最优 or 包含当前元素的局部最优”
但问题来了,local_max[i][j]是啥?没法算啊~
那么,为什么不可以对local_max[i][j]再来一个动态规划求解呢?
于是,有了如下的迭代公式:
local_max[i][j] = max(local_max[i-1][j]+diff, global_max[i-1][j-1]+max(diff,0))
上面的递推公式 把local_max当成寻优目标了,思路还是fellow经典动态规划思路。
但是,有一部分我一开始一直没想通(蓝字部分),按照经典动态规划思路直观来分析,就应该是local_max[i-1][j]啊,怎么还多出来一个diff呢?
leetcode 【 Best Time to Buy and Sell Stock III 】python 实现
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原文地址:http://www.cnblogs.com/xbf9xbf/p/4307840.html