LeetCode OJ 之 Best Time to Buy and Sell Stock II (买卖股票的最佳时间 2)-优快云博客

本文介绍了一种计算股票交易最大利润的算法。该算法允许进行多次买卖操作但不能同时持有多个仓位。通过比较连续两天的价格变化来确定是否进行交易以获得最大累积利润。

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 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).

假设你有一个数组，第i个元素是第i天的股票价格。

设计一个算法，找出最大的利润。你可以完成尽可能多的交易，（即可多次买卖股票）。但是，同一时间不能进行多次交易。

代码：

class Solution {
public:
    int maxProfit(vector<int> &prices) 
    {
        int profit = 0;//总收益
        //思路：求出每相邻两天的收益，收益如果大于0，则加到总收益上，否则就不加
        if(prices.size() < 2)
            return 0;
        for(vector<int>::iterator iter1 = prices.begin() ; iter1 != prices.end()-1 ; iter1++)
        {
            int diff = *(iter1 + 1) - *iter1;//相邻两天的收益
            if(diff > 0)    //收益如果大于0，则加到总收益上
                profit += diff;
        }
        return profit;
    }
};