###### tags: `Leetcode` `medium` `dynamic programming` `greedy` `python`

# 122. Best Time to Buy and Sell Stock II

## [題目連結:] https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/

## 題目:
You are given an integer array ```prices``` where ```prices[i]``` is the price of a given stock on the ```ith``` day.

On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can buy it then immediately sell it on the **same day**.

Find and return the **maximum** profit you can achieve.

**Example 1:**

Input: prices = [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.
Total profit is 4 + 3 = 7.

**Example 2:**

Input: prices = [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.
Total profit is 4.

**Example 3:**

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

## 解題想法:
兄弟題目: [121. Best Time to Buy and Sell Stock](/k1o9uteuT8GbpkFCT0isEw)

* 題目為求買賣最大獲利
* 可以進行多次買賣
* Greedy:
* 因為可以進行多次買賣:
* 可以每天都買，隔天立刻賣掉
* 若當天price比前一天高，則前一天買，當天賣
* 相當於先知模式

## Python:

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        #greedy
        profit=0
        for i in range(1,len(prices)):
            #若當天price比前一天高 則前一天買 當天賣
            if prices[i]>prices[i-1]:
                profit+=(prices[i]-prices[i-1])
        return profit

result = Solution()
prices = [7,1,5,3,6,4]
ans = result.maxProfit(prices)
print(ans)