由此,我们从第1天(下标为0)开始,画出全部的状态转换图,如下: 该图中,每一条从0到n的路径表示一种可能的交易方式,则该状态转换图包含了全部可能的交易方式。这就是解空间,有了它我们才能想办法在其中求一个最大值,接下来就是动态规划的内容了。 二,动态规划 上一段中的状态好理解,但是在程序中如何表示和...
}publicstaticvoidmain(String[] args){ BestTimeToBuyAndSellStock bestTimeToBuyAndSellStock =newBestTimeToBuyAndSellStock(); System.out.println(bestTimeToBuyAndSellStock.maxProfit(newint[]{1,2,3,4,5})); System.out.println(bestTimeToBuyAndSellStock.maxProfit(newint[]{1,-2,3,5,-9})); ...
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; ...
leetcode : Best Time to Buy and Sell Stock 最近看到网站上提到了leetcode网站,用来在线面试算法;就上去看了下,自己也解决了一题,蛮有意思的,偶尔做做算法练练脑。 题目:Best Time to Buy and Sell Stock Say you have an array for which theithelement is the price of a given stock on dayi....
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
121. Best Time to Buy and Sell Stock 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 the maximum profit...
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): """
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。
Best Time to Buy and Sell Stock 相对比较简单的方法是用DP,思路是对于每个i都求出从0到i区间内的最大获益,而对于i+1只需要比较第i+1天的价格和前i天最低价的关系,就可以直接求出0到i+1天区间内的最大获益。也就是说对0到i天的最大获益的计算复杂度是O(1),总体复杂度是O(n)。