96 lines
2.5 KiB
Markdown
96 lines
2.5 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/
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## 题目描述
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```
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Say you have an array for which the ith element is the price of a given stock on day i.
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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).
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Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
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Example 1:
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Input: [7,1,5,3,6,4]
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Output: 7
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Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
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Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
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Example 2:
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Input: [1,2,3,4,5]
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Output: 4
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Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
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Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
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engaging multiple transactions at the same time. You must sell before buying again.
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Example 3:
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Input: [7,6,4,3,1]
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Output: 0
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Explanation: In this case, no transaction is done, i.e. max profit = 0.
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```
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## 思路
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由于我们是想获取到最大的利润，我们的策略应该是低点买入，高点卖出。
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由于题目对于交易次数没有限制，因此只要能够赚钱的机会我们都不应该放过。
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> 如下图，我们只需要求出加粗部分的总和即可
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用图表示的话就是这样：
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## 关键点解析
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- 这类题只要你在心中（或者别的地方）画出上面这种图就很容易解决
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## 代码
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语言支持：JS，Python
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JS Code:
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```js
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/**
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* @param {number[]} prices
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* @return {number}
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*/
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var maxProfit = function(prices) {
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let profit = 0;
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for(let i = 1; i < prices.length; i++) {
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if (prices[i] > prices[i -1]) {
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profit = profit + prices[i] - prices[i - 1];
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}
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}
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return profit;
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};
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```
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Python Code:
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```python
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class Solution:
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def maxProfit(self, prices: 'List[int]') -> int:
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gains = [prices[i] - prices[i-1] for i in range(1, len(prices))
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if prices[i] - prices[i-1] > 0]
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return sum(gains)
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#评论区里都讲这是一道开玩笑的送分题.
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```
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## 相关题目
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- [121.best-time-to-buy-and-sell-stock](./121.best-time-to-buy-and-sell-stock.md)
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- [309.best-time-to-buy-and-sell-stock-with-cooldown](./309.best-time-to-buy-and-sell-stock-with-cooldown.md)
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