来源:力扣(LeetCode)链接:https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-iv 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 思路 可以用递归来算,从根开始,要么买入,要么保持不动。买入后可以进行的操作有卖出,保持不动保持不动后的操作有:买入,保持
//由于数组中存的是buysellpointer,因此需要指向指针的指针做为参数,buysellpointer ** void expandBSPHeap(buysellpointer **heap, int *bsppNumLimit) { //分配两倍的数组空间,每个数组中存放的是buysellpointer buysellpointer* newheap = (buysellpointer*)malloc(*bsppNumLimit *2 * sizeof(buysellpointer...
Design an algorithm to find the maximum profit. You may complete at mostktransactions. Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). Credits: Special thanks to@Freezenfor adding this problem and creating all test cas...
Can you solve this real interview question? Best Time to Buy and Sell Stock IV - You are given an integer array prices where prices[i] is the price of a given stock on the ith day, and an integer k. Find the maximum profit you can achieve. You may compl
输入:k = 2, prices = [2,4,1]输出:2解释:在第 1 天 (股票价格 = 2) 的时候买入,在第 2 天 (股票价格 = 4) 的时候卖出,这笔交易所能获得利润 = 4-2 = 2 。 示例2: 输入:k = 2, prices = [3,2,6,5,0,3]输出:7解释:在第 2 天 (股票价格 = 2) 的时候买入,在第 3 天 (股...
输入:k = 2, prices = [2,4,1] 输出:2 解释:在第1 天 (股票价格 = 2) 的时候买入,在第 2 天 (股票价格 = 4) 的时候卖出,这笔交易所能获得利润 = 4-2 = 2 。 示例2: 输入:k = 2, prices = [3,2,6,5,0,3] 输出:7 解释:在第2 天 (股票价格 = 2) 的时候买入,在第 3 天 ...
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/discuss/108870/Most-consistent-ways-of-dealing-with-the-series-of-stock-problems 读完本文,你可以去力扣拿下如下题目: 买卖股票的最佳时机 买卖股票的最佳时机 II ...
则当k > size / 2时,问题可以转化为:Best Time to Buy and Sell Stock II Python代码: classSolution:# @return an integer as the maximum profitdefmaxProfit(self, k, prices):size = len(prices)ifk > size /2:returnself.quickSolve(size, prices) ...
《Best Time to Buy and Sell Stock II》的方法来求解。所以实际上这道题是之前的二和三的综合体。 AC代码: <span style="font-family:Microsoft YaHei;font-size:12px;">publicclassSolution{publicintmaxProfit(intk,int[]prices){if(prices==null||prices.length==0)return0;if(k>prices.length)//k次...
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)。