题目：

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).

解题思路：

前缀pre[i]处理 0 ~ i 买卖一次最优解，后缀suf[i]处理 i ~ prices.size() - 1 买卖一次最优解。
所有位置pre[i] + suf[i]最大值为答案O(n)。
处理最优解的时候是维护前（后）缀prices最小（大）值，与当前prices做差后和前（后）缀最优解比较取最优，O(n)。
总复杂度O(n)。

代码：

class Solution {
public:int maxProfit(vector<int> &prices) {int min_price = 0x3fffffff;vector<int> pre(prices.size()), suf(prices.size());for (int i = 0; i < prices.size(); i++) {min_price = min(min_price, prices[i]);pre[i] = max(prices[i] - min_price, i ? pre[i - 1] : 0);}int max_price = 0x80000000;for (int i = prices.size() - 1; i >= 0; i--) {max_price = max(max_price, prices[i]);suf[i] = max(max_price - prices[i], i < prices.size() - 2 ? suf[i + 1] : 0);}int ans = 0;for (int i = 0; i < prices.size(); i ++) {ans = max(ans, pre[i] + suf[i]);}return ans;}
};

转载于:https://www.cnblogs.com/dongguangqing/p/3727805.html

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