<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[upperTime的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/upperTime</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; upperTime]]></copyright><item><title><![CDATA[进化算法实现]]></title><link>https://blog.youkuaiyun.com/upperTime/article/details/123579431</link><guid>https://blog.youkuaiyun.com/upperTime/article/details/123579431</guid><author>upperTime</author><pubDate>Fri, 18 Mar 2022 16:54:53 +0800</pubDate><description><![CDATA[```java
package evolutionary;

import java.util.*;

import static java.lang.Math.*;

public class Evolutionary {

    //精度
    static double precision= 0.0001f;
    //种群个数
    static int number=100;
    //位数
    static int bit_number;
    //上界
    static .]]></description><category></category></item><item><title><![CDATA[《Docker实战（第2版）》错误整理 基于docker20.10.10]]></title><link>https://blog.youkuaiyun.com/upperTime/article/details/121579923</link><guid>https://blog.youkuaiyun.com/upperTime/article/details/121579923</guid><author>upperTime</author><pubDate>Sat, 27 Nov 2021 17:06:26 +0800</pubDate><description><![CDATA[4,5节共享挂载点和共享文件（78页） 查看来自主机的日志命令错误

原命令为
cat "$(docker volume inspect --format "{{json .Mountpoint}}" logging-example)"/logA

报错如下
cat: "/var/lib/docker/volumes/logging-example/_data"/logA: No such file or directory

更正为
cat "$(docker volume inspect --form.]]></description><category></category></item><item><title><![CDATA[二叉树中所有距离为 K 的结点]]></title><link>https://blog.youkuaiyun.com/upperTime/article/details/100823977</link><guid>https://blog.youkuaiyun.com/upperTime/article/details/100823977</guid><author>upperTime</author><pubDate>Sat, 14 Sep 2019 11:35:32 +0800</pubDate><description><![CDATA[⼆叉树中所有距离为 K 的结点:https://leetcode-cn.com/problems/all-nodes-distance-k-inbinary-tree/
总体思路： 对于二叉树距离为k的结点的寻找，我们首先应该将所有结点变为可达的，对于此问题的解决，可以用深度搜索的方法建立可达的关系，因为深度搜素和二叉树的前序遍历很是相似，不会改变树的结构。之后利用广度搜索的方法进行第k个结点的查...]]></description><category></category></item><item><title><![CDATA[叶子相似的树]]></title><link>https://blog.youkuaiyun.com/upperTime/article/details/100632945</link><guid>https://blog.youkuaiyun.com/upperTime/article/details/100632945</guid><author>upperTime</author><pubDate>Sun, 08 Sep 2019 17:39:10 +0800</pubDate><description><![CDATA[叶子相似的树:https://leetcode-cn.com/problems/leaf-similar-trees/
总体思路：对二叉树进行遍历并将其叶子节点通过字符串拼接到一起，通过字符串对比输出结果
class Solution {
public  boolean leafSimilar(TreeNode root1, TreeNode root2) {
        String r1=...]]></description><category></category></item><item><title><![CDATA[三⾓形最⼩路径和]]></title><link>https://blog.youkuaiyun.com/upperTime/article/details/100632737</link><guid>https://blog.youkuaiyun.com/upperTime/article/details/100632737</guid><author>upperTime</author><pubDate>Sun, 08 Sep 2019 17:35:12 +0800</pubDate><description><![CDATA[三⾓形最⼩路径和:https://leetcode-cn.com/problems/triangle/
思路：自顶向下，将本级的数字加到下一级的相同位置或者左一位，若下一级有两个选择则进行比较，选择和最小的保存到本级，进行循环，二维数组最后一行则为最后的结果，遍历得到最小的路径值输出
class Solution {
public static int minimumTotal(List&lt;L...]]></description><category></category></item><item><title><![CDATA[Bigram 分词]]></title><link>https://blog.youkuaiyun.com/upperTime/article/details/100592203</link><guid>https://blog.youkuaiyun.com/upperTime/article/details/100592203</guid><author>upperTime</author><pubDate>Sat, 07 Sep 2019 10:41:07 +0800</pubDate><description><![CDATA[Bigram 分词:https://leetcode-cn.com/problems/occurrences-after-bigram/
思路：刚开始自己的想法是首先拼接first与second即string total = first + " “+second+” ";通过find函数对字符串total进行查找，若匹配到则从下一个字符开始到下一个空格为止，将其复制到字符数组中，并将母串中的第一个...]]></description><category></category></item><item><title><![CDATA[⼭⽺拉丁⽂]]></title><link>https://blog.youkuaiyun.com/upperTime/article/details/100591957</link><guid>https://blog.youkuaiyun.com/upperTime/article/details/100591957</guid><author>upperTime</author><pubDate>Sat, 07 Sep 2019 10:22:43 +0800</pubDate><description><![CDATA[⼭⽺拉丁⽂：https://leetcode-cn.com/problems/goat-latin/
总体思路：先将字符串通过空格分割并存入字符数组中，再将字符数组中的各字符串依据题目要求进行变换最后再拼接到一起返回
#pragma once
#include&lt;iostream&gt;
#include&lt;string&gt;
using namespace std;
class Sol...]]></description><category></category></item><item><title><![CDATA[双进程dp问题--探寻宝藏]]></title><link>https://blog.youkuaiyun.com/upperTime/article/details/91383679</link><guid>https://blog.youkuaiyun.com/upperTime/article/details/91383679</guid><author>upperTime</author><pubDate>Mon, 10 Jun 2019 16:36:04 +0800</pubDate><description><![CDATA[最近做课设时，找了网上好多关于关于这个问题的博客，但是都是只有最后的结果与建立的备忘表，都没有去标记到底哪个点是哪条路走过的，因此将对行走过点的标记过程写成了一个函数，具体的思想网上也写的很多，推荐几个




https://blog.youkuaiyun.com/greenhandcgl/article/details/51399222









这个解释的比较详细而且给出了另一种做法dfs，但是...]]></description><category></category></item><item><title><![CDATA[常见线性滤波（第一篇博客）]]></title><link>https://blog.youkuaiyun.com/upperTime/article/details/87919876</link><guid>https://blog.youkuaiyun.com/upperTime/article/details/87919876</guid><author>upperTime</author><pubDate>Mon, 25 Feb 2019 21:35:03 +0800</pubDate><description><![CDATA[#线性滤波器]]></description><category></category></item></channel></rss>