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.

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

須要註意的是，能夠操作兩次買賣。可是第二次買入必須在第一次賣出之後才幹操作。所以思路就是先正序使用貪心計算每一天之前買入賣出可能達到的最大收益，拿數組記錄下來。再逆序計算每一天相應的之後買入賣出可能達到的最大收益，拿數組記錄下來。然後將兩個數組中每一天相應的兩種情況能夠實現的收益之和，得到最大值即為能夠實現的最大收益。code例如以下：

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        if(prices.empty())
            return 0;
        int n = prices.size();
        vector<int> pre(n,0);
        vector<int> post(n,0);
        int curMin = prices[0];
        int curMax = prices[n-1];
        int result=0;
        for(int i=1; i<n; i++)
        {
            curMin = curMin<prices[i] ? curMin : prices[i];
            int diff = prices[i]-curMin;
            pre[i] = pre[i-1]>diff ? pre[i-1] : diff;
        }
        for(int i=n-2; i>=0; i--)
        {
            curMax = curMax>prices[i] ? curMax : prices[i];
            int diff = curMax-prices[i];
            post[i] = post[i+1]>diff ?
 
 post[i+1] : diff;
        }
        for(int i=1; i<n; i++)
        {
            result = result>(pre[i]+post[i]) ?
 result : (pre[i]+post[i]);
        }
        return result;
    }
};

leetcode——Best Time to Buy and Sell Stock III 買賣股票最大收益（AC）