题目1:Best Time to Buy and Sell Stock

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 stock), design an algorithm to find
the maximum profit.

分析:

题意给我们一个数组prices[], 用来表示股票每一天的价格,问我们假设“仅仅多进行一次交易”(即买进之后再卖出), 应该在哪天买进,哪天卖出所获得的利润能达到最大值。

如:prices={1, 3, 4, 10, 1};
那么最大的利润是第一天买进(价格为1),然后第四天卖出(价格为10), 利润最大为9

明确了题意之后,我们来看看能如何解决这道题目。

因为是仅仅能交易一次,并且必须先有“买进”,才干有“卖出”
因此我们仅仅须要数组的最后一位往前扫描。依次得到哪一天的prices是最大的(设为maxPrices),然后在算出和这一天前面的某一天prices[i]的差值 (maxPrices - prices[i]),假设大于最大的利润值,则更新最大利润值maxMoney;

AC代码: 
public class Solution {

    public int maxProfit(int[] prices) {
int size = prices.length;
if (size == 0){
return 0;
}
int maxPrice = prices[size-1];//初始化最大price
int maxMoney = 0;//初始化利润值
for (int i=size-1; i>=0; --i){
maxPrice = maxPrice > prices[i] ? maxPrice : prices[i];//假设第i天的值大于最大price,则更新最大price的值
maxMoney = maxMoney > (maxPrice - prices[i]) ? maxMoney : (maxPrice - prices[i]);//更新最大利润值
}
return maxMoney;
}
}

题目2:Best Time to Buy and Sell Stock II

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). However, you may not engage
in multiple transactions at the same time
(ie, you must sell the stock before you buy again).


分析:

这道题目。关键是要理解题意,题意中和题目I是有点像,唯一不同的是这道题目能够进行多次交易,可是要注意的是,每次最多仅仅能持有一支股票在手上,也就是说你要再买入的时候。必须先把手头上的这股票先卖掉!

举个样例:如Prices[] = {1,3,4,10,2};
这样子的话,你能够採取的方式有
1、第1天买入(price==1)。第2天卖出(price==3), 第3天买入(price==4),第4天卖出(price==10)  :   8
2、第1天买入(price==1)。第3天卖出(price==4)       : 3
3、第2天买入(price==3),第3天卖出(price==4)       : 1
4、第3天买入(price==4),第4天卖出(price==10)     : 6
5、第1天买入(price==1),第4天卖出(price==10)     : 9
但事实上假设细致观察easy发现 : 
3 - 1 = 2
4 - 3 = 1
10 - 4 = 6

然后result = 2 + 1 + 6 = 9
因此事实上我们仅仅是要找增长的序列对。并求出他们的差值的和
index :  1 ~~ prices.size()-1
通过这样分析的话,我们非常easy知道事实上仅仅要从头到尾遍历,假设 prices[index] > prices[index-1] 

AC代码:

public class Solution {
public int maxProfit(int[] prices) {
int profit = 0;
int size = prices.length;
if (size < 2){
return profit;
}
for (int index=1; index<size; ++index){
int value = prices[index] - prices[index-1];
if (value > 0){
profit += value;
}
}
return profit;
}
}

题目3:

Best Time to Buy and Sell Stock III

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 may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

分析:
因为最多仅仅能够进行两次交易,并且两次交易之间必须没有交集。
这题用DP来做.
那我们非常easy想到。将数组划分成两块,左边一块 | 右边一块,我们仅仅须要求出左边的最大利润,再求出右边的最大利润。然后相加起来就得到了利润的最大值
我们用
用left[i] 来表示[0,...,i]中的最大利润

用right[i]来表示[i,...,n-1]上的最大利润

AC代码:
public class Solution {
public int maxProfit(int[] prices) {
int size = prices.length;
if (size < 2)
return 0;
int[] left = new int[size];
int[] right = new int[size];
int minValue = prices[0];
int maxValue = prices[size-1];
for (int i=1; i<size; ++i){
left[i] = left[i-1] > (prices[i] - minValue) ? left[i-1] : (prices[i] - minValue);
minValue = minValue < prices[i] ? minValue : prices[i];
}
for (int i=size-2; i>=0; --i){
right[i] = right[i+1] > (maxValue - prices[i]) ? right[i+1] : (maxValue - prices[i]);
maxValue = maxValue > prices[i] ? maxValue : prices[i];
}
int profit=0;
for (int i=0; i<size; ++i){
profit = profit > (left[i] + right[i]) ? profit : (left[i] + right[i]);
}
return profit;
}
}


Best Time to Buy and Sell Stock I &amp;&amp; II &amp;&amp; III的更多相关文章

  1. [LeetCode] 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 al ...

  2. [LeetCode] 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 al ...

  3. [LeetCode] 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 al ...

  4. [LeetCode] 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 al ...

  5. [LeetCode] 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 ...

  6. [LintCode] 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 al ...

  7. [LintCode] 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 ...

  8. LeetCode——Best Time to Buy and Sell Stock II (股票买卖时机问题2)

    问题: Say you have an array for which the ith element is the price of a given stock on day i. Design a ...

  9. 123. Best Time to Buy and Sell Stock (三) leetcode解题笔记

    123. Best Time to Buy and Sell Stock III Say you have an array for which the ith element is the pric ...

随机推荐

  1. 2013-8-6 10:56:07 JAVA_WEB:员工号自动生成源代码

    create table user_info_temp (       usId varchar2(20),       usNo varchar2(20),       usName varchar ...

  2. [Node.js]Express web框架

    摘要 Express是一个简洁灵活的node.js web应用框架,提供了一系列强大特性帮助你创建各种web应用和丰富的http工具.使用express可以快速创建一个完整功能的网站. Express ...

  3. Inside Portable Class Libraries

    Portable Class Libraries were introduced with Visual Studio 2010 SP1 to aid writing libraries that c ...

  4. 使用jquery加载部分视图01-使用$.get()

    使用Html.RenderParital或Html.RenderAction可以在主视图中加载部分视图. 两种方法是有区别的,在"RenderPartial和RenderAction区别&q ...

  5. python文本 字符串开头或者结尾匹配

    python文本 字符串开头或者结尾匹配 场景: 字符串开头或者结尾匹配,一般是使用在匹配文件类型或者url 一般使用startwith或者endwith >>> a='http:/ ...

  6. 使用 MVC 5 的 EF6 Code First 入门 系列:建立一个EF数据模型

    这是微软官方SignalR 2.0教程Getting Started with Entity Framework 6 Code First using MVC 5 系列的翻译,这里是第一篇:建立一个E ...

  7. ngx_lua实现登录逻辑

    最近在公司做一个简单的portal,本来很简单的,只用ngx_lua就可以实现所有的业务逻辑,不需要upstream上游服务.但被要求接入公司内部的用户校验系统,说白了就是一个登录过程,只允许公司内部 ...

  8. 第七章 JVM性能监控与故障处理工具(1)

    1.定位系统问题 依据 GC日志 堆转储快照(heapdump/hprof文件) 线程快照(threaddump/javacore文件) 运行日志 异常堆栈 分析依据的工具 jps:显示指定系统内的所 ...

  9. 附 Java对象内存布局

    注意:本篇博客,主要参考自<深入理解Java虚拟机(第二版)> 1.对象在内存中存储的布局分为三块 对象头 存储对象自身的运行时数据:Mark Word(在32bit和64bit虚拟机上长 ...

  10. jquery选择器的实现流程简析及提高性能建议!

    当我们洋洋得意的使用jquery强大的选择器功能时有没有在意过jquery的选择性能问题呢,其实要想高效的使用jquery选择器,了解其实现流程是很有必要的,那么这篇文章我就简单的讲讲其实现流程,相信 ...