【leetcode】Best Time to Buy and Sell Stock III

Best Time to Buy and Sell Stock III

Say you have 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 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).

 

设dpBack[i]为从第一天到第i天的最大的收益

设dpAfter[j]为从第j天到最后一天的最大收益

 

如何求最大收益可以考虑采用Best Time to Buy and Sell Stock I的方法

 

注意是最多买两次，也可以只买一次

 

 class Solution {
 public:
     int maxProfit(vector<int> &prices) {

         int n=prices.size();
         if(n==)return ;

         vector<int> dpBack(n),dpAfter(n);

         int maxProfit=;
         dpBack[]=;
         int left=prices[];

         for(int i=;i<n;i++)
         {
             if(prices[i]>left)
             {
                if(maxProfit<prices[i]-left) maxProfit=prices[i]-left;
             }
             else
             {
                 left=prices[i];
             }

             dpBack[i]=maxProfit;
         }

         maxProfit=;
         dpAfter[n-]=;
         int right=prices[n-];

         for(int j=n-;j>=;j--)
         {
             if(prices[j]<right)
             {
                 if(maxProfit<right-prices[j]) maxProfit=right-prices[j];
             }
             else
             {
                 right=prices[j];
             }

             dpAfter[j]=maxProfit;
         }

         int result=;
         for(int i=;i<n-;i++)
         {
             if(dpBack[i]+dpAfter[i+]>result) result=dpBack[i]+dpAfter[i+];

             if(result<dpBack[i+]) result=dpBack[i+];
         }

         return result;
     }
 };