C++实现LeetCode(309.买股票的最佳时间含冷冻期)

[LeetCode] 309.Best Time to Buy and Sell Stock with Cooldown 买股票的最佳时间含冷冻期

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) with the following restrictions:

  • You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
  • After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)

Example:

prices = [1, 2, 3, 0, 2]
maxProfit = 3
transactions = [buy, sell, cooldown, buy, sell]

这道题又是关于买卖股票的问题，之前有四道类似的题目Best Time to Buy and Sell Stock 买卖股票的最佳时间，Best Time to Buy and Sell Stock II 买股票的最佳时间之二， Best Time to Buy and Sell Stock III 买股票的最佳时间之三和Best Time to Buy and Sell Stock IV 买卖股票的最佳时间之四。而这道题与上面这些不同之处在于加入了一个冷冻期Cooldown之说，就是如果某天卖了股票，那么第二天不能买股票，有一天的冷冻期。根据他的解法，此题需要维护三个一维数组buy, sell，和rest。其中：

buy[i]表示在第i天之前最后一个操作是买，此时的最大收益。

sell[i]表示在第i天之前最后一个操作是卖，此时的最大收益。

rest[i]表示在第i天之前最后一个操作是冷冻期，此时的最大收益。

我们写出递推式为：

buy[i]  = max(rest[i-1] - price, buy[i-1])
sell[i] = max(buy[i-1] + price, sell[i-1])
rest[i] = max(sell[i-1], buy[i-1], rest[i-1])

上述递推式很好的表示了在买之前有冷冻期，买之前要卖掉之前的股票。一个小技巧是如何保证[buy, rest, buy]的情况不会出现，这是由于buy[i] <= rest[i]， 即rest[i] = max(sell[i-1], rest[i-1])，这保证了[buy, rest, buy]不会出现。

另外，由于冷冻期的存在，我们可以得出rest[i] = sell[i-1]，这样，我们可以将上面三个递推式精简到两个：

buy[i]  = max(sell[i-2] - price, buy[i-1])
sell[i] = max(buy[i-1] + price, sell[i-1])

我们还可以做进一步优化，由于i只依赖于i-1和i-2，所以我们可以在O(1)的空间复杂度完成算法，参见代码如下：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int buy = INT_MIN, pre_buy = 0, sell = 0, pre_sell = 0;
        for (int price : prices) {
            pre_buy = buy;
            buy = max(pre_sell - price, pre_buy);
            pre_sell = sell;
            sell = max(pre_buy + price, pre_sell);
        }
        return sell;
    }
};

类似题目：

Best Time to Buy and Sell Stock IV

Best Time to Buy and Sell Stock III

Best Time to Buy and Sell Stock II

Best Time to Buy and Sell Stock

参考资料：

https://leetcode.com/discuss/71354/share-my-thinking-process

到此这篇关于C++实现LeetCode(309.买股票的最佳时间含冷冻期)的文章就介绍到这了,更多相关C++实现买股票的最佳时间含冷冻期内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们！

赞 (0)

相关推荐

  • C++实现LeetCode(557.翻转字符串中的单词之三)

    [LeetCode] 557.Reverse Words in a String III 翻转字符串中的单词之三 Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order. Example 1: Input: "Let's take LeetCode conte

  • C++实现LeetCode(188.买卖股票的最佳时间之四)

    [LeetCode] 188.Best Time to Buy and Sell Stock IV 买卖股票的最佳时间之四 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 k transactions. Note: You m

  • C++实现LeetCode(169.求大多数)

    [LeetCode] 169. Majority Element 求大多数 Given an array nums of size n, return the majority element. The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array. Example 1

  • C++实现LeetCode(170.两数之和之三 - 数据结构设计)

    [LeetCode] 170. Two Sum III - Data structure design 两数之和之三 - 数据结构设计 Design and implement a TwoSum class. It should support the following operations: add and find. add - Add the number to an internal data structure. find - Find if there exists any pai

  • C++实现LeetCode(186.翻转字符串中的单词之二)

    [LeetCode] 186. Reverse Words in a String II 翻转字符串中的单词之二 Given an input string , reverse the string word by word.  Example: Input:  ["t","h","e"," ","s","k","y"," ","i&qu

  • C++实现LeetCode(173.二叉搜索树迭代器)

    [LeetCode] 173.Binary Search Tree Iterator 二叉搜索树迭代器 Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note: next() and

  • C++实现LeetCode(172.求阶乘末尾零的个数)

    [LeetCode] 172. Factorial Trailing Zeroes 求阶乘末尾零的个数 Given an integer n, return the number of trailing zeroes in n!. Example 1: Input: 3 Output: 0 Explanation: 3! = 6, no trailing zero. Example 2: Input: 5 Output: 1 Explanation: 5! = 120, one trailing

  • C++实现LeetCode(309.买股票的最佳时间含冷冻期)

    [LeetCode] 309.Best Time to Buy and Sell Stock with Cooldown 买股票的最佳时间含冷冻期 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

  • C++实现LeetCode(123.买股票的最佳时间之三)

    [LeetCode] 123.Best Time to Buy and Sell Stock III 买股票的最佳时间之三 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 two transactions. Note: You

  • C++实现LeetCode(122.买股票的最佳时间之二)

    [LeetCode] 122.Best Time to Buy and Sell Stock II 买股票的最佳时间之二 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

  • C++实现LeetCode(121.买卖股票的最佳时间)

    [LeetCode] 121.Best Time to Buy and Sell Stock 买卖股票的最佳时间 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 sto

  • Java通过动态规划设计股票买卖最佳时机

    Java通过动态规划设计股票买卖最佳时机

    目录 买卖股票的最佳时机 动态规划 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大利润.如果你不能获取任何利润,返回 0 . 示例: 链接:https://leetcode.cn/problems/best-time-to-buy-and-sell-stock 动态规划

  • JavaScript股票的动态买卖规划实例分析下篇

    目录 1. 最佳买卖股票时机含冷冻期 题目描述 题解 2. 买卖股票的最佳时机 III 题目描述 题解 1. 最佳买卖股票时机含冷冻期 题目描述 给定一个整数数组prices,其中第prices[i]表示第i天的股票价格 . 设计一个算法计算出最大利润.在满足以下约束条件下,你可以尽可能地完成更多的交易(多次买卖一支股票): 卖出股票后,你无法在第二天买入股票 (即冷冻期为 1 天). 注意:你不能同时参与多笔交易(你必须在再次购买前出售掉之前的股票). 示例 1: 输入: prices = [

  • 用js实现每隔一秒刷新时间的实例(含年月日时分秒)

    用js实现每隔一秒刷新时间的实例(含年月日时分秒)

    原理:使用定时器,即setInterval(fn,i),每隔i秒执行fn. 下面给出具体的代码 1.代码如下: <span style="font-size:14px;"><!DOCTYPE html> <html lang="zh-cn"> <head> <meta charset="UTF-8"> <title>用js实现每隔一秒刷新时间(含年月日时分秒)</tit

  • 如何让SQL运行得更快

    一.不合理的索引设计   ----例:表record有620000行,试看在不同的索引下,下面几个 SQL的运行情况:   ---- 1.在date上建有一非个群集索引   select count(*) from record where date >   '19991201' and date < '19991214'and amount >   2000 (25秒)   select date,sum(amount) from record group by date   (55秒

  • [转载]让SQL运行得更快

    如何让你的SQL运行得更快     人们在使用SQL时往往会陷入一个误区,即太关注于所得的结果是否正确,而忽略了不同的实现方法之间可能存在的性能差异,这种性能差异在大型的或是复杂的数据库环境中(如联机事务处理OLTP或决策支持系统DSS)中表现得尤为明显.笔者在工作实践中发现,不良的SQL往往来自于不恰当的索引设计.不充份的连接条件和不可优化的where子句.在对它们进行适当的优化后,其运行速度有了明显地提高!下面我将从这三个方面分别进行总结:   ----为了更直观地说明问题,所有实例中的SQ

随机推荐