題目

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).

題目大意：

假設(shè)你有一個數(shù)組，其中第i 個元素是第i天給定股票的價格。
設(shè)計一個算法來找到最大的利潤。您可以根據(jù)需要完成盡可能多的交易（即，購買一次并多次出售股票）。但是，您可能不會同時從事多個交易（即，您必須在再次購買之前先出售該股票）。

解題思路

最低點買入，最高點賣出，所以只需要計算增長過程中相關(guān)聯(lián)的最低點到最高點差值即可

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