Leetcode "best time to Buy and Sell Stock III" Python implementation

title :

Say you has an array for which the i-th element is the price of a given-stock on day I.

Design an algorithm to find the maximum profit. You are in most of the transactions.

Note:
Engage in multiple transactions on the same time (ie, you must sell the stock before you buy again).

Code : runtime:175 ms

1 classSolution:2     #@param prices, a list of integers3     #@return An integer4     defmaxprofit_with_k_transactions (self, prices, K):5Days =len (Prices)6Local_max = [[0 forIinchRange (k+1)] forIinchrange (days)]7Global_max = [[0 forIinchRange (k+1)] forIinchrange (days)]8          forIinchRange (1, days):9diff = prices[i]-prices[i-1]Ten              forJinchRange (1,k+1): OneLOCAL_MAX[I][J] = max (Local_max[i-1][j]+diff, global_max[i-1][j-1]+Max (diff,0)) AGLOBAL_MAX[I][J] = max (Local_max[i][j], global_max[i-1][j]) -         returnGlobal_max[days-1][k] -  the     defMaxprofit (Self, prices): -         ifPrices isNoneorLen (Prices) <2: -             return0 -         returnSelf.maxprofit_with_k_transactions (Prices, 2)

Ideas :

Not your own thinking, refer to this blog http://blog.csdn.net/fightforyourdream/article/details/14503469

The same as the above blog thinking is not repeated, the following is their own experience:

1. This type of topic, the ultimate train of thought must be to the dynamic planning, I myself summed up as " global optimal = all elements before the current element of the optimal or contains the current element of the optimal "

2. The difficulty with this problem is that the optimization cannot be accomplished by an iterative formula.

Ideas are as follows:

GLOBAL_MAX[I][J] = max (Global_max[i-1][j], local_max[i][j])

The idea of the iterative formula above is clear: "The global Optimal to the first element = The global optimal or contains the local optimal" of the current element, which does not contain the first element.

But here's the question, local_max[i][j] what is it? It can't be counted.

So why not have a dynamic solver for LOCAL_MAX[I][J]?

As a result, the following iteration formula is available:

LOCAL_MAX[I][J] = max (local_max[i-1][j]+diff, Global_max[i-1][j-1]+max (diff,0))

The above recursive formula regards Local_max as the goal of optimization, and the idea is to fellow the classical dynamic planning.

However, there is a part I have not figured out (blue word part), according to the classic dynamic planning ideas intuitive to analyze, it should be local_max[i-1][j] ah, how to come out a diff it?

Leetcode "best time to Buy and Sell Stock III" Python implementation