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...
【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 ...
第三题:https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/?tab=Description 在第二题的基础上加入了一个cooldown。 还是沿用上一题的思路,sell之后的第一天不能立刻buy,而是必须等一天才可以buy。 那么上面的方程就要做一个变换: buy[i] = max(sell[i - 2]...
【leetcode】41-best-time-to-buy-and-sell-stock-ii 力扣 122. 买卖股票的最佳时机 II 【leetcode】42-best-time-to-buy-and-sell-stock-iii 力扣 123. 买卖股票的最佳时机 III 【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-bu...
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 the maximum profit. code : 朴素的算法,对每一个i ,计算 最大 的 prices[j] (j>i) 来维护最大的差值,但是这样的复杂度是O(n^2),会TLE ...
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 - 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
Can you solve this real interview question? Best Time to Buy and Sell Stock III - You are given an array prices where prices[i] is the price of a given stock on the ith day. Find the maximum profit you can achieve. You may complete at most two transacti
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
int buy=prices[0]; int sell = 0; int day= 0; while(day<prices.size()){ if(prices[day]<buy) { buy =prices[day]; } sell = max(prices[day]-buy,sell); day++; } return sell; } }; 展开全部 2 回复 dp小亡子 来自 北京 ...