# 【LeetCode】 122. 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 (i.e., buy one and sell one share of the stock multiple times).

> Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

> 給你一個陣列，每個元素代表第i天的物價。

> 設計一個演算法找到最大的利潤，你應該會完成許多次的價差交易(買和賣很多次，但一次的亮只能買賣一個)。

> 注意：你不應該一次進行多個交易(你必須先買才賣)。

## Example:

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
             
             
Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.
             
             
Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

## Solution

* 只需要在意前一次和下一次的價格是否為正，然後加總上去。

### Code

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size()<2)return 0;
        int ans=0;
        for(int i=0;i<prices.size()-1;i++)
        {
            ans+=max(0,prices[i+1]-prices[i]);
        }
        return ans;
    }
};

###### tags: `LeetCode` `C++`