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:## 对于任意一个右指针日期,我们都要回顾它后面的所有日期,
所以,对于第i天而言,可以选择不卖,那么以卖为最后一动作的最大利润就等于sell[i - 1],也可以选择在第i天卖,那么就是prices[i]加上到第i-1天为止的以买为最后动作的最大利润,即prices[i] + buy[i - 1]。 所以sell[i] = max(sell[i - 1], buy[i - 1] + prices[i])。 buy[i]:第i天为...
【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】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 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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit. ...
[LeetCode] Best Time to Buy and Sell Stock 买卖股票的最佳时间,Sayyouhaveanarrayforwhichtheithelementisthepriceofagivenstockondayi.Ifyouwereonlypermittedtocompleteatmostonetransaction(ie,buyoneandselloneshareofthestock),des
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
Input:k = 2, prices = [3,2,6,5,0,3]Output:7Explanation:Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. ...
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 2025.04.19 保证一个最小值,有比他小的就更新 ...
Problem: 思路 设置一个元素 buy 代表买入价格, sell 代表卖出价格。(sell价格一定是恒大于等于buy) 对 prices 进行遍历,如果当前价格高于 sell ,则更新 sell Python3 贪心 1 61 0夏天的风园游会 ・ 2020.02.03 经典DP class Solution { public: int maxProfit(vector<int>& prices) { if(prices.size()...