best time to Buy and Sell Stock III

best time to Buy and Sell Stock IIITotal accepted:30820 Total submissions:130535my submissions

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).

1#include <vector>2#include <iostream>3 using namespacestd;4 5 6 intMain ()7 {8     intprisize, tem,i;9vector<int>pri;Ten      while(Cin >>prisize) One     { A pri.clear (); -          for(i =0; i < prisize; ++i) -         { theCIN >>tem; - Pri.push_back (TEM); -         } -  +         if(prisize<=1)  -         { +cout <<0<<Endl; A             Continue; at         } -         if(prisize==2)   -         { -             intTEM = pri[1]>pri[0]?pri[1]-pri[0]:0; -cout << tem <<Endl; -             Continue; in         } -  to  +         //in the first case, I bought it only once. -         //1, from left to right traverse save min Price min, Judge Pri[i]-min > maxprice?, update the current maximum profit Maxprice the         //2, from left to right traverse save Max Price Max, Judge Max-pri[i] > maxprice?, update Maxprice *  $         //in the second case, I bought it two times.Panax Notoginseng         //3, max (d) = Maxprice (0 <= D <= i) + maxprice (I < D < prisize) -         //D traverse between 0 < D < priSize-1, interpret Max (d) > Maxprice, update maxprice the         //results of Maxprice (0 <= D <= i) and Maxprice (I < D < prisize) +         //can be recorded in the Maxfromleft,maxfromright in the 1th and 2 steps in advance. A         //4, maxfromleft[prisize-1] = = maxfromright[0] = = The maximum profit to buy only once the         //use one of them to compare the size with the 3rd, Maxprice, and get the result +  -  $         //algorithmic complexity O (n-1 + n-1 + n-2) = O (3n-4) = O (n) $  -vector<int> Maxfromleft (prisize,0);  -vector<int> Maxfromright (prisize,0); the         intMIN = Int_max, max=int_min, Maxprice =Int_min,pritem; -          for(intI=0; i<prisize;i++)Wuyi         { the             //Update the minimum price starting from the left -             if(pri[i]<MIN) WuMIN =Pri[i]; -Pritem = pri[i]-MIN; About  $             //Update current maximum profit -             if(Pritem >maxprice) -Maxprice =Pritem; -  A             //Save Maxprice (0 <= D <= i) +Maxfromleft[i] =Maxprice; the         } -  $Maxprice =int_min; the          for(inti=prisize-1; i>=0;--i) the         { the             //Update the maximum price starting from the right the             if(pri[i]>MAX) -MAX =Pri[i]; inPritem = MAX-Pri[i]; the             //Update current maximum profit the             if(Pritem >maxprice) AboutMaxprice =Pritem; the             //Save Maxprice (i < D < prisize) theMaxfromright[i] =Maxprice; the         } +  -Maxprice =int_min; the          for(intI=0; i<prisize-1; i++){Bayi             //Update current maximum profit thePritem = maxfromleft[i]+maxfromright[i+1]; the             if(Pritem > Maxprice) Maxprice =Pritem; -         } -  the         //maxfromleft[prisize-1] and maxprice compare size, Draw results the         if(maxfromleft[prisize-1] >maxprice) theMaxprice = maxfromright[0]; the  -cout << Maxprice <<Endl; the     } the  the     return 0 ;94}

best time to Buy and Sell Stock III