right = 0, 1 # left=buy, right=sell maxP = 0 while right < len(prices): ## 遍历整个 list if prices[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
【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】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-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-...
publicintmaxProfit(int[] prices){if(prices ==null|| prices.length <=1) {return0; }intmaxProfit = Integer.MIN_VALUE;intbuyPrice = prices[0];for(inti=1; i < prices.length; i++) {if(prices[i] > buyPrice) { maxProfit = Math.max(maxProfit, prices[i]-buyPrice); }else{ buyPrice ...
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/ 这里f1[i],要这样理解,第i天是可以卖出的(包括第i天),肯定不是买入。for a particular i, 我们可以按照 I 题的办法,求得当前价格减去当前min的最大值求得f1[i]的值,但是我们如果对于不同的i都从头到尾scan一遍求得当前价格减去当...
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
121. 买卖股票的最佳时机 - 给定一个数组 prices ,它的第 i 个元素 prices[i] 表示一支给定股票第 i 天的价格。 你只能选择 某一天 买入这只股票,并选择在 未来的某一个不同的日子 卖出该股票。设计一个算法来计算你所能获取的最大利润。 返回你可以从这笔交易中获取的
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class Solution { public: int maxProfit(vector<int>& prices) { int n = prices.size(); int dp[n][2]; //vector 更耗时 memset(dp, 0, sizeof(dp)); int minv = prices[0], maxv = prices[n - 1], ans = 0...