股票买卖2次

Example 1:

Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

class Solution {

public:
    int maxProfit(vector& prices) {
        
        int size = prices.size();
        if(size == 0){
            return 0;
        }
        // 买卖2两次，dpl[i] 从左边开始的0到i天的最大利润值 
        vector dpl(size,0);
        int lprice=prices[0];
        for(int i=1;i            if(lprice>prices[i]){
                lprice=prices[i]; // 0-i 最小值
            }
            dpl[i] = max(dpl[i-1],prices[i]-lprice);
        }
        
        // dpr[i] 从右边开始的  i 到 size 直接的最大值利润值
        vector dpr(size,0);
        int rprice = prices[size-1];
        for(int i=size-2;i>=0;--i){
            if(rprice                rprice=prices[i]; // i-size 最大值
            }
            dpr[i] = max(dpr[i+1],rprice-prices[i]);
        }
        
        int res=0;
        for(int i=0;i            res=max(res,(dpl[i]+dpr[i]));
        }
        
        return res;
    }
};