Say you have an array for which theithelement 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 the maximum profit. 大意是:有一个数组,数组的下标表示第几...
代码: 1publicintmaxProfit(int[] prices) {2intlen =prices.length ;3if(len < 2)return0;45intmin = prices[0] ;6intmaxProfit = 0;78for(inti = 1 ; i < len ; i++){9inttemp = prices[i] - min;//当前值减去前i-1个值的最小值10if(maxProfit < temp) maxProfit = temp;//更新最大...
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
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 the stock), design an algorithm to find the maximum profit. 思路分析: 给定一个数组,数组中第i...
LeetCode Best Time to Buy and Sell Stock 买卖股票的最佳时机 (DP),题意:给定一个序列,第i个元素代表第i天这支股票的价格,问在最佳时机买入和卖出能赚多少钱?只买一次,且仅1股,假设本钱无限。思路:要找一个最低价的时候买入,在最高价的时候卖出利润会最大。
LeetCode 0121. Best Time to Buy and Sell Stock买卖股票的最佳时机【Easy】【Python】【贪心】 Problem 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...
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
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...