122.Best Time to Buy and Sell Stock II
===

###### tags: `Medium`,`Array`,`DP`,`Greedy`

[122. Best Time to Buy and Sell Stock II](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/)

### 題目描述

You are given an array `prices` where `prices[i]` is the price of a given stock on the i^th^ 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*.

### 範例

**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.

**Constraints**:

* 1 <= `prices.length` <= 10^5^
* 0 <= `prices[i]` <= 10^4^

### 解答

#### Javascript

function maxProfit(prices) {
  let max = 0;
  let minPrice = prices[0];
  for (let i = 1; i < prices.length; i++) {
    if (prices[i] < minPrice) {
      minPrice = prices[i];
    } else if (prices[i] > minPrice) {
      max += prices[i] - minPrice;
      minPrice = prices[i];
    }
  }

  return max;
}

>[name=Marsgoat][time= Dec 23, 2022]

#### Python

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        pre, ans = math.inf, 0
        for price in prices:
            ans += max(0, price - pre)
            pre = price
        return ans

> [name=Yen-Chi Chen][time=Fri, Dec 23, 2022]

### Reference

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