[leetcode] best time to buy and sell stocks 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).

不是一个很typical的思路，因为只能有2次transaction，所以。。。从前扫一次，从后面扫一次，分别存在2个buffer里面，然后在扫一次buffer就好啦

class Solution {
public:

    int maxProfit(vector<int> &prices) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(prices.empty()) return 0;

        int minVal = prices[0];
        int result = 0;

        vector<int> inOrder;
        inOrder.push_back(0);
        vector<int> revOrder;
        revOrder.assign(prices.size(),0);

        for(int i = 1; i < prices.size(); i++){
            if(prices[i] > minVal){
                result = MAX(result, ( prices[i] - minVal));
            }else{
                minVal = prices[i];
            }
            inOrder.push_back(result);
        }

        int maxVal = prices[prices.size()-1];
        result = 0;

        for(int i = prices.size()-2; i >= 0; i--){
            if(prices[i] < maxVal){
                result = MAX(result, maxVal - prices[i] );
            }else{
                maxVal = prices[i];
            }
            revOrder[i] = result;
        }


        result = 0;
        for (int i = 0; i < inOrder.size()-1; i++) {
            result = MAX(result, inOrder[i] + revOrder[i+1]);
        }

        result = MAX(result, revOrder[0]);

        return result;

    }

};