LeetCode-188.Best Time to Buy and Sell Stock IV
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 must sell the stock before you buy again).
Example 1:
Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
Example 2:
Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. 使用动态规划
public int maxProfit(int k, int[] prices) {//dp mytip
if(null==prices||0==prices.length||0>=k){
return 0;
}
int max =0;
if(k>prices.length){//若k过大 优化
for (int i = 0; i < prices.length-1; i++) {
if(prices[i]<prices[i+1]){
max+= prices[i+1]-prices[i];
}
}
}
else{
int[][][] states = new int[prices.length][k+1][2];//状态表示第i个数在第j此交易中,有无股票时(0为无,1为有)的利益;//因为只保存上一个数时的利益,所以states可优化为[k+1][2]
for(int i=0;i<=k;i++){//初始化第1个数的状态
states[0][i][1]=-prices[0];
}
for(int i=1;i<prices.length;i++){
for(int j=0;j<=k;j++){
if(j==0){
states[i][j][0] = states[i-1][j][0];//防止j-1溢出
}
else{
states[i][j][0] = Math.max(states[i-1][j][0],states[i-1][j-1][1]+prices[i]);
}
states[i][j][1] = Math.max(states[i-1][j][1],states[i-1][j][0]-prices[i]);
}
}
for(int i=0;i<=k;i++){
max = max>states[prices.length-1][i][0]?max:states[prices.length-1][i][0];
}
} return max;
}
优化空间
public int maxProfit(int k, int[] prices) {//dp my
if(null==prices||0==prices.length||0>=k){
return 0;
}
int max =0;
if(k>prices.length){
for (int i = 0; i < prices.length-1; i++) {
if(prices[i]<prices[i+1]){
max+= prices[i+1]-prices[i];
}
}
}
else{
int[][] states = new int[k+1][2];
states[0][1] = -prices[0];
for(int i=0;i<=k;i++){
states[i][1]=-prices[0];
}
for(int i=1;i<prices.length;i++){
for(int j=0;j<=k;j++){
states[j][1] = Math.max(states[j][1],states[j][0]-prices[i]);
if(j==0){
states[j][0] = states[j][0];
}
else{
states[j][0] = Math.max(states[j][0],states[j-1][1]+prices[i]);
} }
}
for(int i=0;i<=k;i++){
max = max>states[i][0]?max:states[i][0];
}
} return max;
}
相关题
买卖股票的最佳时间1 LeetCode121 https://www.cnblogs.com/zhacai/p/10429264.html
买卖股票的最佳时间2 LeetCode122 https://www.cnblogs.com/zhacai/p/10596627.html
买卖股票的最佳时间3 LeetCode123 https://www.cnblogs.com/zhacai/p/10645571.html
买卖股票的最佳时间冷冻期 LeetCode309 https://www.cnblogs.com/zhacai/p/10655970.html
买卖股票的最佳时间交易费 LeetCode714 https://www.cnblogs.com/zhacai/p/10659288.html
LeetCode-188.Best Time to Buy and Sell Stock IV的更多相关文章
- Java for LeetCode 188 Best Time to Buy and Sell Stock IV【HARD】
Say you have an array for which the ith element is the price of a given stock on day i. Design an al ...
- [LeetCode] 188. Best Time to Buy and Sell Stock IV 买卖股票的最佳时间 IV
Say you have an array for which the ith element is the price of a given stock on day i. Design an al ...
- LeetCode 188. Best Time to Buy and Sell Stock IV (stock problem)
Say you have an array for which the ith element is the price of a given stock on day i. Design an al ...
- 【刷题-LeetCode】188 Best Time to Buy and Sell Stock IV
Best Time to Buy and Sell Stock IV Say you have an array for which the i-th element is the price of ...
- 【LeetCode】Best Time to Buy and Sell Stock IV
Best Time to Buy and Sell Stock IV Say you have an array for which the ith element is the price of a ...
- 【LeetCode】188. Best Time to Buy and Sell Stock IV 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...
- [LeetCode][Java] Best Time to Buy and Sell Stock IV
题目: Say you have an array for which the ith element is the price of a given stock on day i. Design a ...
- 188. Best Time to Buy and Sell Stock IV (Array; DP)
Say you have an array for which the ith element is the price of a given stock on day i. Design an al ...
- 188. Best Time to Buy and Sell Stock IV leetcode解题笔记
Say you have an array for which the ith element is the price of a given stock on day i. Design an al ...
- 188. Best Time to Buy and Sell Stock IV——LeetCode
Say you have an array for which the ith element is the price of a given stock on day i. Design an al ...
随机推荐
- (实用)CentOS 6.3更新内置Python2.6
在安装Kilo版的OpenStack时,我们发现社区已经将Python升到2.7,而CentOS 6.3上仍然在使用2.6版的Python.本文记录将CentOS 6.3内置的Python2.6更新为 ...
- 跨浏览器的placeholder-jQuery版(jQuery插件EnPlaceholder)
案例:整搜索框,需要默认占位符为"请输入关键词",获取焦点时,占位符消失或不可用(不影响正常输入),丢失焦点后,若用户无内容输入,占位符继续出现,继续占位.这种代码我想前端们已经很 ...
- orcal数据库基本操作
1.连接 SQL*Plus system/manager 2.显示当前连接用户 SQL> show user 3.查看系统拥有哪些用户 SQL> select * from all_use ...
- MangoDB学习笔记
01. 数据库操作 1. 查看当前数据库名称 db 2. 查看所有数据库名称,列出所有在物理上存在的数据库 show dbs; 3. 切换数据库,如果数据库不存在也并不创建,直到插入数据或创建集合时数 ...
- linux erase
map的erase windows和linux不同,而迭代器弄不好就失效 1 #include <iostream> 2 #include <map> 3 #include & ...
- [IOI 2000]POJ 1160 Post Office
Post Office Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 22278 Accepted: 12034 Descrip ...
- NLP入门资料
<统计自然语言处理> 一些基础理论概念,涉及统计自然语言处理的基本概念.理论方法和新研究进展,内容包括形式语言与自动机及其在自然语言处理中的应用.语言模型.隐马尔可夫模型.语料库技术.汉语 ...
- Docker 修改国内镜像地址
curl -sSL https://get.daocloud.io/daotools/set_mirror.sh | sh -s http://86d2a50b.m.daocloud.io 该脚本可以 ...
- ARM v8中断机制和中断处理(转)
https://blog.csdn.net/firefox_1980/article/details/40113637 https://blog.csdn.net/firefox_1980/artic ...
- ASP.NET异步
1.ASP.NET线程模型 在WEB程序中,天生就是多线程的,我们知道,一个WEB服务可以同时服务器多个用户,我们可以想象一下,WEB程序应该运行于多线程环境中,对于运行WEB程序的线程,我们可以称之 ...