# 122. Best Time to Buy and Sell Stock II

**<font color=red>难度:Easy<font>**

## 刷题内容
> 原题连接

* https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/

> 内容描述

```
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).

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.
```

> 思路
******- 时间复杂度: O(N)******- 空间复杂度: O(1)******

用暴力的解法很快就可以做出来这题，但是我们可以用更好的方法在O(N)的时间复杂度解决这题。定义i = 1。遍历数组。如果 prices[i] > prices[i - 1]。ans += （prices[i] - prices[i - 1]）。

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