1/**2* Design an algorithm to find the maximum profit. You may complete as many transactions as you like3* (ie, buy one and sell one share of the stock multiple times).4* However, you may not engage in multiple transactions at the same time5* (ie, you must sell the stock before ...
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 ...
Can you solve this real interview question? Best Time to Buy and Sell Stock II - 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 h
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ 因为可以进行无限次交易,并且在下一次buy之前必须已经sell。所以只需要把所有price曲线价格上涨的部分加起来就行。 class Solution(object): def maxProfit(self, prices): """ :type prices: List[int] :rtype: int """ maxprofit = 0...
Leetcode: Best Time to Buy and Sell Stock II 题目: 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 (ie, buy one and sell one share of the ...
第0天时,buy1=−prices[i],buy2=−prices[i]当天买入又卖出,再买入,sell1=sell2=0 最终的结果为sell2, 因为sell2包含了一次交易和二次交易的最大收益 classSolution{public:intmaxProfit(vector<int>&prices){intn=prices.size();intdp[n][2];//vector 更耗时memset(dp,0...
122. 买卖股票的最佳时机 II - 给你一个整数数组 prices ,其中 prices[i] 表示某支股票第 i 天的价格。 在每一天,你可以决定是否购买和/或出售股票。你在任何时候 最多 只能持有 一股 股票。你也可以先购买,然后在 同一天 出售。 返回 你能获得的 最大 利润 。 示
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)。
遍历价格vector,把每段上升期间的利润加起来,注意最后结束时记得再计算最后一次的利润 class Solution { // 把每段上升期间的利润加起来 // minVal: 低谷值 // maxVal: 峰值 // profit: 利润 public: int maxProfit(vector
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...