題目：

Say you have an array for which the i^th 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).

解題思路：

這題與上一題Best Time to Buy and Sell Stock相比，其實還更簡單，掃描一遍陣列，當前前元素大於前一個元素，則將其差值累加到max中，最後求得的max即為最大利益。

程式碼：

#include <iostream>
#include <vector>

using namespace std;

/**
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).

 
*/

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

int main(void)
{
    int arr[] = {2,4,5,1,7,10};
    int n = sizeof(arr) / sizeof(arr[0]);
    vector<int> stock(arr, arr+n);
    Solution solution;
    int max = solution.maxProfit(stock);
    cout<<max<<endl;
    return 0;
    return 0;
}