Runtime: 68 ms, faster than 5.85% of Python3 online submissions for Best Time to Buy and Sell Stock II. Memory Usage: 14.2 MB, less than 5.06% of Python3 online submissions for Best Time to Buy and Sell Stock II. 所以想提高,还是要学一下迭代器的应用啊。 标签: python , LeetCode ...
Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). ...
LeetCode 0122. Best Time to Buy and Sell Stock II买卖股票的最佳时机 II【Easy】【Python】【贪心】【动态规划】 Problem LeetCode Say you have an array for which theith element is the price of a given stock on dayi. Design an algorithm to find the maximum profit. You may complete as man...
01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第32题(顺位题号是122)。假设有一个数组,其中第i个元素是第i天给定股票的价格。设计算法以找到最大利润。可以根据需要完成尽可能多的交易(即,多次买入并卖出一股股票)。 注意:不能同时进行多笔交易(即,您必须在再次购买之前卖出股票)。 例如: 输入:[7,...
【LeetCode】【Python解决问题的方法】Best Time to Buy and Sell Stock II,Sayyouhaveanarrayforwhichtheithelementisthepriceofagivenstockonday
Leetcode Best Time to Buy and Sell Stock II Say you have an array for which theith element is the price of a given stock on dayi. Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock ...
classSolution:defmaxProfit(self,prices:List[int])->int:## 双指针解法left,right=0,1# left=buy, right=sellmaxP=0whileright<len(prices):## 遍历整个 listifprices[right]>prices[left]:## 在存在赚钱机会的条件下profit=prices[right]-prices[left]maxP=max(maxP,profit)else:## 对于任意一个右指...
1.Violence to solve:Basically, if tomorrow's price is higher than today's, we buy it today and sell tomorrow. Otherwise, we don't. Here is the python_code: classSolution(object):defmaxProfit(self,prices):# Through the arrayforiinrange(len(prices)-1):max_price=sum(max(prices[i+1]-...
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