<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[JunyuLee的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/JunyuLee</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; JunyuLee]]></copyright><item><title><![CDATA[详解Huffman编码 - Java实现]]></title><link>https://blog.youkuaiyun.com/JunyuLee/article/details/106761433</link><guid>https://blog.youkuaiyun.com/JunyuLee/article/details/106761433</guid><author>JunyuLee</author><pubDate>Mon, 15 Jun 2020 14:17:27 +0800</pubDate><description><![CDATA[关于Huffman编码的具体定义可参考《算法导论》P245

Huffman Tree节点定义:
abstract class AbstractNode {
    protected int weight;

    protected abstract AbstractNode setWeight(int weight);

    protected abstract int getWeight();

    protected AbstractNode(int weight) {
       .]]></description><category></category></item><item><title><![CDATA[详解dp - 最长回文子序列]]></title><link>https://blog.youkuaiyun.com/JunyuLee/article/details/106752549</link><guid>https://blog.youkuaiyun.com/JunyuLee/article/details/106752549</guid><author>JunyuLee</author><pubDate>Sun, 14 Jun 2020 21:59:18 +0800</pubDate><description><![CDATA[给定一个字符串 s ，找到其中最长的回文子序列，并返回该序列的长度。可以假设 s 的最大长度为 1000 。
类似问题 : 最长回文子串
首先找出状态转移方程 : table[i][j] = table[i+1][j-1] + 2, if s[i] == s[j] ; else table[i][j] = max(table[i+1][j], table[i][j-1])，同最长回文子串的求解方法进行求解
code :
class Solution {
    public int longestPalin]]></description><category></category></item><item><title><![CDATA[详解 dp - 最长回文字串]]></title><link>https://blog.youkuaiyun.com/JunyuLee/article/details/106752338</link><guid>https://blog.youkuaiyun.com/JunyuLee/article/details/106752338</guid><author>JunyuLee</author><pubDate>Sun, 14 Jun 2020 21:53:22 +0800</pubDate><description><![CDATA[给定一个字符串 s，找到 s 中最长的回文子串。你可以假设 s 的最大长度为 1000。
思路 : 先给出状态转移方程，然后分别由备忘递归或者dp求解；本题中，对于任意的两个合法下标 i -
j，table[i][j] = table[i+1][j-1] &amp;&amp; (s[i] == s[j]) ， 考虑边缘情况，由于这个表达式只能适用于长度 &gt;= 3的字符串，考虑长度 == 1的字符串，table[i][i] = true，考虑长度 == 2 的字符串 : table[i][j] = s[]]></description><category></category></item><item><title><![CDATA[钢条切割问题 - Java实现（递归，备忘录递归，动态规划）]]></title><link>https://blog.youkuaiyun.com/JunyuLee/article/details/106752207</link><guid>https://blog.youkuaiyun.com/JunyuLee/article/details/106752207</guid><author>JunyuLee</author><pubDate>Sun, 14 Jun 2020 21:35:04 +0800</pubDate><description><![CDATA[因为这个问题比较简单，就不给详细解答，直接贴代码了
public class RodCut {
    public static void main(String[] args) {
        int [] price = {0,1,5, 8,9,10,17,17,20,24,30,34,38,43,47,50,57,62,63,64,65,66,67,68,69,70, 71, 82};
        final int num = price.length - 1;
        int[]]]></description><category></category></item><item><title><![CDATA[最优二叉搜索树 - Java实现（递归，带备忘录递归，动态规划）]]></title><link>https://blog.youkuaiyun.com/JunyuLee/article/details/105959148</link><guid>https://blog.youkuaiyun.com/JunyuLee/article/details/105959148</guid><author>JunyuLee</author><pubDate>Wed, 06 May 2020 21:33:38 +0800</pubDate><description><![CDATA[对于一个给定的概率集合，我们希望构造一棵期望搜索代价最小的二叉搜索树，我们称之为最优二叉搜索树

最近复习了一下算法导论的动态规划，发现最优二叉搜索树的实现方法与矩阵链连乘问题极其相似。
根据动态规划bottom -&gt; top的思想，我们应该从length = 1的情况逐渐求解到length = n的情况，对于每一个i = length，我们从起始位置 j = 1开始，加上length -...]]></description><category></category></item><item><title><![CDATA[详解矩阵链乘法问题 - Java实现版本（递归与动态规划）]]></title><link>https://blog.youkuaiyun.com/JunyuLee/article/details/105671719</link><guid>https://blog.youkuaiyun.com/JunyuLee/article/details/105671719</guid><author>JunyuLee</author><pubDate>Wed, 22 Apr 2020 01:24:59 +0800</pubDate><description><![CDATA[矩阵链乘法问题：给定n个矩阵的链&lt;A1,A2,…,An&gt;，矩阵Ai的规模为pi-1*pi(1&lt;=i&lt;=n)，求完全括号化方案，使得计算乘积A1A2…An所需标量乘法次数最少

看到这种具有最优子结构的问题，第一时间就应该想到使用动态规划的思想来解决，假设dp[i][j]是Ai…Aj的最小代价括号化子问题，当i==j时是同一矩阵不需要相乘，所以dp[i][i] = 0 ( ...]]></description><category></category></item><item><title><![CDATA[从源码入门Future与FutureTask的使用]]></title><link>https://blog.youkuaiyun.com/JunyuLee/article/details/105642683</link><guid>https://blog.youkuaiyun.com/JunyuLee/article/details/105642683</guid><author>JunyuLee</author><pubDate>Tue, 21 Apr 2020 00:37:38 +0800</pubDate><description><![CDATA[Future
话不多说，直接看一下文档对于Future接口的解释

重点:Future是

]]></description><category></category></item></channel></rss>