最多两次股票交易-Best Time to Buy and Sell Stock III

Say you have an array for which the ith 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.（股票交易，最多两次）

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        if(prices.size()==0)return 0;
        int n=prices.size();
        vector<int> left(n);
        vector<int> right(n);
        int min=prices[0];
        int max=prices[n-1];
        int res=0;
        for(int i=1;i<n;i++)
        {
            min=min<prices[i]?min:prices[i];
            left[i]=left[i-1]>(prices[i]-min)?left[i-1]:(prices[i]-min);
        }
        for(int j=n-2;j>=0;j--)
        {
            max=max>prices[j]?max:prices[j];
            right[j]=right[j+1]>(max-prices[j])?right[j+1]:(max-prices[j]);
        }
        for(int i=0;i<n;i++)
        {
            res=res>(left[i]+right[i])?res:(left[i]+right[i]);
        }
        return res;
    }
};