股票交易

题目描述

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).

定义两个数组，before从前向后遍历，第i个位置代表前i天买卖一次的最大收益
after从后向前遍历，第i个位置代表i天之后买卖一次的最大收益
最后从前向后遍历一遍，找出before[i]和after[i]和的最大值即可

class Solution {
public:
   int maxProfit(vector<int> &prices)
{
    if(prices.size()<=1)
        return 0;
    int sz=prices.size();
    vector<int> before(sz,0);
    int min_val=prices[0];
    int max_mon=0;
    for(int i=1;iif(prices[i]-min_val>max_mon)
        {   
            max_mon=prices[i]-min_val;
        }   
        if(prices[i]1]);
    }   
    int max_val=prices[sz-1];
    max_mon=0;
    vector<int> after(sz,0);
    for(int i=sz-2;i>=0;--i)
    {   
        if(max_val-prices[i]>max_mon)
            max_mon=max_val-prices[i];
        if(max_val1],max_val-prices[i]);
    }   
    int maxsum=0;
    for(int i=0;iif(before[i]+after[i]>maxsum)
            maxsum=before[i]+after[i];
    }
    return maxsum;
}

};