【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==0)return 0;
 
        vector<int> dpBack(n),dpAfter(n);
 
        int maxProfit=0;
        dpBack[0]=0;
        int left=prices[0];
       
        for(int i=1;i<n;i++)
        {
            if(prices[i]>left)
            {
               if(maxProfit<prices[i]-left) maxProfit=prices[i]-left;
            }
            else
            {
                left=prices[i];
            }
           
            dpBack[i]=maxProfit;
        }
       
        maxProfit=0;
        dpAfter[n-1]=0;
        int right=prices[n-1];
       
        for(int j=n-2;j>=0;j--)
        {
            if(prices[j]<right)
            {
                if(maxProfit<right-prices[j]) maxProfit=right-prices[j];
            }
            else
            {
                right=prices[j];
            }
 
            dpAfter[j]=maxProfit;
        }
       
        int result=0;
        for(int i=0;i<n-1;i++)
        {
            if(dpBack[i]+dpAfter[i+1]>result) result=dpBack[i]+dpAfter[i+1];
           
            if(result<dpBack[i+1]) result=dpBack[i+1];
        }
       
        return result;
    }
};