Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) wit…

使用动态规划,下面的代码可以通过210个测试,最后1个(第211个)会超时.说明思路是正确的,但是其中有一些是无效的计算. class Solution { public: int maxProfit(vector<int>& prices) { int n=prices.size(); ) { ; } ; int **D; D = new int*[N]; ;i<n;i++){ D[i]=new int[N]; } ;i<n;i++){ for(int j=i;j<…

一.穷举框架 首先,还是一样的思路:如何穷举?这里的穷举思路和上篇文章递归的思想不太一样. 递归其实是符合我们思考的逻辑的,一步步推进,遇到无法解决的就丢给递归,一不小心就做出来了,可读性还很好.缺点就是一旦出错,你也不容易找到错误出现的原因.比如上篇文章的递归解法,肯定还有计算冗余,但确实不容易找到. 而这里,我们不用递归思想进行穷举,而是利用「状态」进行穷举.我们具体到每一天,看看总共有几种可能的「状态」,再找出每个「状态」对应的「选择」.我们要穷举所有「状态」,穷举的目的是根据对应的「选择…

Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer feerepresenting a transaction fee. You may complete as many transactions as you like, but you need to pay the tr…

Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete at most ktransactions. Note:You may not engage in multiple transactions at the same time (ie, you…

继承leetcode123以及leetcode309的思路,,但应该也可以写成leetcode 152. 乘积最大子序列的形式 class Solution { public: int maxProfit(vector<int>& prices, int fee) { int len=prices.size(); ) ; vector<]-fee); vector<); ; ;i<len;i++){ int p=prices[i]; buy[i]=max(buy[i-…