<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[LiLi_code的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/LiLi_code</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; LiLi_code]]></copyright><item><title><![CDATA[react 跨域解决]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/128291721</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/128291721</guid><author>LiLi_code</author><pubDate>Mon, 12 Dec 2022 19:05:50 +0800</pubDate><description><![CDATA[react配置跨域代理方法 & localhost拒绝访问的问题]]></description><category></category></item><item><title><![CDATA[yarn start 运行react项目失败]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/125568542</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/125568542</guid><author>LiLi_code</author><pubDate>Sat, 02 Jul 2022 07:53:47 +0800</pubDate><description><![CDATA[开发中的爬坑之路，react运行]]></description><category></category></item><item><title><![CDATA[信号处理中的低通、高通、带通、带阻滤波器]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/123752911</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/123752911</guid><author>LiLi_code</author><pubDate>Sat, 26 Mar 2022 11:31:48 +0800</pubDate><description><![CDATA[低通滤波器
	
允许低于截止频率的信号通过，但高于截止频率的信号不能通过的电子滤波装置。




	高通滤波器
	
  允许高于截止频率的信号通过。是一种让某一频率以上的信号分量通过，而对该频率以下的信号分量大大抑制的电容、电感与电阻等器件的组合装置。其特性在时域及频域中可分别用冲激响应及频率响应描述。




	带通滤波器
	
  允许特定频率的信号通过。是指能通过某一频率范围内的频率分量、但将其他范围的频率分量衰减到极低水平的滤波器，与带阻滤波器的概念相对。一个模拟带通滤波...]]></description><category></category></item><item><title><![CDATA[注释Vue中所有的console.log]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/119749618</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/119749618</guid><author>LiLi_code</author><pubDate>Tue, 17 Aug 2021 09:58:00 +0800</pubDate><description><![CDATA[【一键注释】

在前端开发的过程中一般会打印很多的日志但是要发布到线上的时候不希望客户看到其他的信息，但是一个一个的注释太麻烦也不利于以后进行更新
下面是个很简单的方法一行代码解决
在main.js中加入以下一行代码，全局生效


console.log = function() {}

如果要进行修改代码进行打印日志查看的时候就把这行代码注释就OK~
...]]></description><category></category></item><item><title><![CDATA[PC端移动端适配项目，可以拿来做我的作品集吗哈哈]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/119617526</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/119617526</guid><author>LiLi_code</author><pubDate>Wed, 11 Aug 2021 21:26:20 +0800</pubDate><description><![CDATA[http://www.iclabs.cn/#/pc

之前的实习开发项目~
]]></description><category></category></item><item><title><![CDATA[类比工厂，通俗易懂的讲解进程与线程之间的关系]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/119531476</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/119531476</guid><author>LiLi_code</author><pubDate>Mon, 09 Aug 2021 10:04:29 +0800</pubDate><description><![CDATA[1.计算机的核心是CPU，它承担了所有的计算任务。它就像一座工厂，时刻在运行。



2.假定工厂的电力有限，一次只能供给一个车间使用。也就是说，一个车间开工的时候，其他车间都必须停工。背后的含义就是，单个CPU一次只能运行一个任务。



3.进程就好比工厂的车间，它代表CPU所能处理的单个任务。任一时刻，CPU总是运行一个进程，其他进程处于非运行状态。



4.一个车间里，可以有很多工人。他们协同完成一个任务。



5.线程就好比车间里的工人。一个进程可以包括多个线程。



6...]]></description><category></category></item><item><title><![CDATA[Vue2.0 $set()]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/119221417</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/119221417</guid><author>LiLi_code</author><pubDate>Thu, 29 Jul 2021 21:53:23 +0800</pubDate><description><![CDATA[最近工作中遇到了这种我额外往对象中添加了属性的情况，导致页面效果一直无法实现。困惑中突然明白了原因，做一下整理总结。


vue2.0 给data对象新增属性，并触发视图更新

如下代码，给 student对象新增 age 属性


data () {
    return {
        student: {
            name: '',
            sex: ''
        }
    }
}

直接给student赋值操作，虽然可以新增属性，但是不会触发视图更新
.]]></description><category></category></item><item><title><![CDATA[JavaScript刷题时的输入问题总结]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118381372</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118381372</guid><author>LiLi_code</author><pubDate>Thu, 01 Jul 2021 10:49:34 +0800</pubDate><description><![CDATA[牛客编译模式选择node编译器：

单行输入：

题目链接：https://www.nowcoder.com/practice/171278d170c64d998ab342b3b40171bb?tpId=40&amp;tqId=21336&amp;tPage=1&amp;rp=1&amp;ru=/ta/kaoyan&amp;qru=/ta/kaoyan/question-ranking


const readline = require('readline');
const r1 = readline.c]]></description><category></category></item><item><title><![CDATA[webpack 自定义loader]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118358719</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118358719</guid><author>LiLi_code</author><pubDate>Wed, 30 Jun 2021 13:07:26 +0800</pubDate><description><![CDATA[main.js 文件


function A() {
    console.log('hello world');
}

webpack.config.js 文件


const path = require('path'); 

module.exports = {
    entry: path.join(__dirname, './main.js'), 
    output: {
        path: path.resolve(__dirname, 'dist'),
        fil]]></description><category></category></item><item><title><![CDATA[LeetCode912. 排序数组（JavaScript手撕各种排序算法）]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118309943</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118309943</guid><author>LiLi_code</author><pubDate>Tue, 29 Jun 2021 09:41:16 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/sort-an-array/submissions/

这是一道好题目，可以用来练习手撕各种排序算法，别直接调用api错过了这道好题哦！

目录

一、插入排序

直接插入排序：超时

折半插入排序：AC

希尔排序：AC

二、交换排序

冒泡排序：AC

快速排序：AC

三、选择排序

选择排序：AC

堆排序：AC

四、归并排序

归并排序：AC

五、JavaScript 内部 API

直接调用API：AC

一、..]]></description><category></category></item><item><title><![CDATA[Vue中DOM异步更新的解决方案——$nextTick]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118309487</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118309487</guid><author>LiLi_code</author><pubDate>Mon, 28 Jun 2021 18:23:10 +0800</pubDate><description><![CDATA[html:


&lt;body&gt;
&lt;div id="app"&gt;
    &lt;audio ref="audio" :src="url" controls&gt;&lt;/audio&gt;
    &lt;span&gt;第 {{index}} 首歌&lt;/span&gt;
    &lt;button @click="changeUrl"&gt;切歌&lt;/button&gt;
&lt;/div&gt;
&lt;/body&gt;

js:


&lt;script src="h]]></description><category></category></item><item><title><![CDATA[LeetCode46. 全排列（JavaScript解析）]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118280161</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118280161</guid><author>LiLi_code</author><pubDate>Sun, 27 Jun 2021 21:21:44 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/permutations/submissions/

回溯大法好啊！ 我还用了标记数组，回溯完记得恢复原状


/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var permute = function(nums) {
    let ans = [] //存放所有的排列结果
    let temp = [] //存放每一次的排列数据
    let len =]]></description><category></category></item><item><title><![CDATA[LeetCode349. 两个数组的交集]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118253057</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118253057</guid><author>LiLi_code</author><pubDate>Sat, 26 Jun 2021 18:50:34 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/intersection-of-two-arrays/

使用set求2个数组的交集：


var intersection = function(nums1, nums2) {
    var set = new Set(nums1)
    //Array.from 就是将一个类数组对象或者可遍历对象转换成一个真正的数组
    nums2 = Array.from(new Set(nums2)) //为nums2去重
    ]]></description><category></category></item><item><title><![CDATA[LeetCode160.相交链表]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118251644</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118251644</guid><author>LiLi_code</author><pubDate>Sat, 26 Jun 2021 17:00:49 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/intersection-of-two-linked-lists/

方法一：

非常的巧妙优美，可以让两个节点走完自己的路再去走对方的路，从而走的节点数会是一样长的


var getIntersectionNode = function(headA, headB) {
    if(!headA || !headB) return null;
    let A = headA
    let B = headB
    whi]]></description><category></category></item><item><title><![CDATA[LeetCode121. 买卖股票的最佳时机（JavaScript解析）]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118251131</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118251131</guid><author>LiLi_code</author><pubDate>Sat, 26 Jun 2021 16:29:45 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/

DP思想：

1. 记录【今天之前买入的最小值】，是为了在最小值买

2. 计算【今天之前最小值买入，可以获得的利润】，这便是今天卖出的话的最大利润

3. 比较【每天的最大利润】，取最大值


/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(price]]></description><category></category></item><item><title><![CDATA[LeetCode94. 二叉树的中序遍历（JavaScript解析）]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118222216</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118222216</guid><author>LiLi_code</author><pubDate>Fri, 25 Jun 2021 16:11:51 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/binary-tree-inorder-traversal/

同样有递归法与非递归法



递归法：


/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefine]]></description><category></category></item><item><title><![CDATA[LeetCode704. 二分查找（JavaScript解析）]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118221155</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118221155</guid><author>LiLi_code</author><pubDate>Fri, 25 Jun 2021 15:03:21 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/binary-search/

就是一道经典的二分查找题目，可以使用递归，也可以使用非递归



递归解析：


/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
    let left = 0;
    let right = nums.leng]]></description><category></category></item><item><title><![CDATA[LeetCode206. 反转链表（JavaScript解析）]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118220318</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118220318</guid><author>LiLi_code</author><pubDate>Fri, 25 Jun 2021 14:29:40 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/reverse-linked-list/


/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @para]]></description><category></category></item><item><title><![CDATA[LeetCode53. 最大子序和 (JavaScript 4种方法解析)]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118106578</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118106578</guid><author>LiLi_code</author><pubDate>Tue, 22 Jun 2021 17:05:26 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/maximum-subarray/

经典题就需要认真做，我主要用三种方法完成本题

方法一：暴力


/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function(nums) {
    if(nums.length == 1){
        return nums[0];
    }
    let ans = -99999;]]></description><category></category></item><item><title><![CDATA[LeetCode165. 比较版本号 (JavaScript解析)]]></title><link>https://blog.youkuaiyun.com/LiLi_code/article/details/118094826</link><guid>https://blog.youkuaiyun.com/LiLi_code/article/details/118094826</guid><author>LiLi_code</author><pubDate>Mon, 21 Jun 2021 21:13:52 +0800</pubDate><description><![CDATA[题目链接：https://leetcode-cn.com/problems/compare-version-numbers/


/**
 * @param {string} version1
 * @param {string} version2
 * @return {number}
 */
var compareVersion = function(version1, version2) {

    let arr1 = version1.split('.');
    let arr2 = ver]]></description><category></category></item></channel></rss>