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leetcode 【 Best Time to Buy and Sell Stock III 】python 实现

时间:2015-03-02 00:55:58      阅读:237      评论:0      收藏:0      [点我收藏+]

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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

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