【leetcode】43-best-time-to-buy-and-sell-stock-iv 力扣 188. 买卖股票的最佳时机 IV 【leetcode】44-best-time-to-buy-and-sell-stock-with-cooldown 力扣 309. 买卖股票的最佳时机包含冷冻期 【leetcode】45-best-time-to-buy-and-sell-stock-with-cooldown 力扣 714. 买卖股票的最佳时机包含手续费 ...
【leetcode】Best Time to Buy and Sell (easy) 题目: Say you have an array for which theith element is the price of a given stock on dayi. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find th...
Best Time to Buy and Sell Stock Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of...
[LeetCode] Best Time to Buy and Sell Stock II 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 a......
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): """
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
思路分析: 这道题《Leetcode: Best Time to Buy and Sell Stock 》类似。不过这次不限制交易次数,所以每次只要是盈利的,我们就进行买入和卖出。 C++参考代码: class Solution { public: int maxProfit(vector<int> &prices) { vector<int>::size_type size = prices.size(); ...
Can you solve this real interview question? Best Time to Buy and Sell Stock - You are given an array prices where prices[i] is the price of a given stock on the ith day. You want to maximize your profit by choosing a single day to buy one stock and choo
https://leetcode.com/problems/best-time-to-buy-and-sell-stock/leetcode.com/problems/best-time-to-buy-and-sell-stock/ 解题思路 1. 暴力 O(n2) 复杂度,超时 class Solution { public: int maxProfit(vector<int>& prices) { int maxVal = 0; int n = prices.size(); for(int i = 0; ...
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...