"Leetcode" best time to Buy and Sell Stock III

best time to Buy and Sell Stock III

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).

Dpback[i] for the maximum benefit from the first day to the day I dpafter[j] for the maximum benefit from the J Day to the last day how to maximize revenue consider the method of using best time to buy and Sell Stock I note that you can buy up to two times or only once

1 classSolution {2  Public:3     intMaxprofit (vector<int> &prices) {4        5         intn=prices.size ();6         if(n==0)return 0;7  8vector<int>dpback (n), Dpafter (n);9  Ten         intmaxprofit=0; Onedpback[0]=0; A         intleft=prices[0]; -         -          for(intI=1; i<n;i++) the         { -             if(prices[i]>Left ) -             { -                if(Maxprofit<prices[i]-left) maxprofit=prices[i]-Left ; +             } -             Else +             { Aleft=Prices[i]; at             } -             -dpback[i]=Maxprofit; -         } -         -maxprofit=0; indpafter[n-1]=0; -         intright=prices[n-1]; to         +          for(intj=n-2; j>=0; j--) -         { the             if(prices[j]<Right ) *             { $                 if(Maxprofit<right-prices[j]) maxprofit=right-Prices[j];Panax Notoginseng             } -             Else the             { +right=Prices[j]; A             } the   +dpafter[j]=Maxprofit; -         } $         $         intresult=0; -          for(intI=0; i<n-1; i++) -         { the             if(dpback[i]+dpafter[i+1]>result) result=dpback[i]+dpafter[i+1]; -            Wuyi             if(result<dpback[i+1]) result=dpback[i+1]; the         } -         Wu         returnresult; -     } About};

"Leetcode" best time to Buy and Sell Stock III