LeetCode--122. Best Time to Buy and Sell Stock II-优快云博客

本文介绍了一种算法，用于解决股票交易中如何通过多次买卖实现最大收益的问题。该算法通过比较连续两天的价格差来决定是否进行交易，从而累积最大利润。

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Best Time to Buy and Sell Stock II

Description：

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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Solution：

考虑的是买进卖出利益最大化问题。

假设序列为：a <= b <= c <= d，那么最大利润为：d-a = （d-c）+（c-b）+（b-a）；

若序列为：a <= b >= c <= d <= e，那么最大利润为：（b-a）+（e-c）=（b-a）+（0）+（d-c）+（e-d）；

所以，maxProfit = max{price[I]- price[I-1]，0}

Code：

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int pro = 0;
        for (int i = 1; i < prices.size(); i++) 
            pro += max(prices[i] - prices[i - 1], 0);    
        return pro;
    }
};