中山大学算法课程题目详解（第十周）

问题描述：

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).

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Seen this question in a real interview before?   

Yes

 

解决思路：

分析：此题亦即变形，也就是求交易次数最多为两次。当然有两种方法，第一种暴力，对于每个i，我们求[0,i]与[i,n-1]两次收益，然后求和，遍历i，可以取其中的最大值，需要O(N^2)的时间。第二种方法是动态规划，用两个数组，第一个数组f1[i]用来表示在[0,i]内进行买入卖出的最大收益，用f2[i]表示在[i,n-1]内进行买入卖出的最大收益。然后最大收益即为max(f1[i]+f2[i])，如何求f1[i]和f2[i]，这里给出第二种方法的代码

代码实现：

int maxProfit(vector<int>& prices) {if (prices.size() <= 1) return 0;int n = prices.size();vector<int> part1(n + 1, 0);vector<int> part2(n + 1, 0);int minPrice = prices[0];for (int i = 1; i < n; i++) {minPrice = min(minPrice, prices[i]);part1[i] = max(part1[i - 1], prices[i] - minPrice);}int maxPrice = prices[n - 1];for (int i = n - 2; i >= 0; i--) {maxPrice = max(maxPrice, prices[i]);part2[i] = max(part2[i + 1], maxPrice - prices[i]);}int maxResult = 0;for (int i = 0; i < n; i++) {maxResult = max(maxResult, part1[i] + part2[i]);}return maxResult;}