<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[perseverance_wang的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/perseverance_wang</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; perseverance_wang]]></copyright><item><title><![CDATA[pandas之多重影分身]]></title><link>https://blog.youkuaiyun.com/perseverance_wang/article/details/127118101</link><guid>https://blog.youkuaiyun.com/perseverance_wang/article/details/127118101</guid><author>perseverance_wang</author><pubDate>Fri, 30 Sep 2022 09:43:26 +0800</pubDate><description><![CDATA[pandas之多重影分身]]></description><category></category></item><item><title><![CDATA[将日期类型转换为int类型]]></title><link>https://blog.youkuaiyun.com/perseverance_wang/article/details/127084322</link><guid>https://blog.youkuaiyun.com/perseverance_wang/article/details/127084322</guid><author>perseverance_wang</author><pubDate>Wed, 28 Sep 2022 10:19:00 +0800</pubDate><description><![CDATA[将日期类型转换为int类型]]></description><category></category></item><item><title><![CDATA[排序]]></title><link>https://blog.youkuaiyun.com/perseverance_wang/article/details/105470033</link><guid>https://blog.youkuaiyun.com/perseverance_wang/article/details/105470033</guid><author>perseverance_wang</author><pubDate>Tue, 14 Apr 2020 20:25:23 +0800</pubDate><description><![CDATA[冒泡排序 / Bubble Sort
插入排序 / Insertion Sort
归并排序 / Merge Sort
快速排序 / Quick Sort
拓扑排序 / Topological Sort
堆排序 / Heap Sort
通排序 / Bucket Sort

]]></description><category></category></item><item><title><![CDATA[区块连简介]]></title><link>https://blog.youkuaiyun.com/perseverance_wang/article/details/105519397</link><guid>https://blog.youkuaiyun.com/perseverance_wang/article/details/105519397</guid><author>perseverance_wang</author><pubDate>Tue, 14 Apr 2020 20:12:09 +0800</pubDate><description><![CDATA[区块连技术概述
技术起源
	1.P2P网络
	2.加密
	3.数据库技术
	4.电子现金

区块连1.0
	1.分布式账本
	2.块链式数据
	3.默克尔树
	4.工作量证明

区块连2.0
	1.智能合约
	2.虚拟机
	3.去中心化应用

区块连3.0
	1.社会治理
	2.价值交换

区块连并不是一项全新的技术，它是对P2P网络、加解密、分布式存储、共识算法等多项成熟技术的一种创新型组合应用...]]></description><category></category></item><item><title><![CDATA[高级数据结构]]></title><link>https://blog.youkuaiyun.com/perseverance_wang/article/details/105448737</link><guid>https://blog.youkuaiyun.com/perseverance_wang/article/details/105448737</guid><author>perseverance_wang</author><pubDate>Sat, 11 Apr 2020 12:37:16 +0800</pubDate><description><![CDATA[优先队列 / Priority Queue
与普通队列的区别
	1.保证每次取出的元素时队列中优先级最高的
	2.优先级别可自定义
场景
	从杂乱无序的数据中按照一定的顺序（或者优先级）筛选数据

本质
	1.二叉堆的结构，堆在英文里叫Binary Heep 
	2.利用一个数组结构来实现完全二叉树

特性
	1.数组里的第一个元素array[0]拥有最高的优先级
	2.给定一个下标i，那么对于元...]]></description><category></category></item><item><title><![CDATA[常用数据结构]]></title><link>https://blog.youkuaiyun.com/perseverance_wang/article/details/105440473</link><guid>https://blog.youkuaiyun.com/perseverance_wang/article/details/105440473</guid><author>perseverance_wang</author><pubDate>Fri, 10 Apr 2020 20:43:18 +0800</pubDate><description><![CDATA[常用数据结构常用数据结构1.数组、字符串 / Array &amp; String2.链表 / Linked-list3.栈 / Stask4.队列5.双端队列 / Deque6.树 / Tree
常用数据结构
1.数组、字符串 / Array &amp; String
数组：
	优点：
		1.构建一个数组非常简单
		2.能让我们在O(1)的时间里根据数组下标(index)查询某个元素
	缺点...]]></description><category></category></item><item><title><![CDATA[Leetcode-08-String to Integer (atoi)]]></title><link>https://blog.youkuaiyun.com/perseverance_wang/article/details/89004902</link><guid>https://blog.youkuaiyun.com/perseverance_wang/article/details/89004902</guid><author>perseverance_wang</author><pubDate>Wed, 03 Apr 2019 21:02:41 +0800</pubDate><description><![CDATA[Leetcode-08-String to Integer (atoi)
题目链接
解题思路：atoi的实现，有很多细节需要注意。两个判断条件很关键。INT_MAX/10 &gt;= sum和INT_MAX - digit &gt;= sum这两个判断条件确保了不会溢出。
class Solution:
    # @return an integer
    def atoi(self, str...]]></description><category></category></item><item><title><![CDATA[Leetcode-07 Reverse Integer]]></title><link>https://blog.youkuaiyun.com/perseverance_wang/article/details/88981055</link><guid>https://blog.youkuaiyun.com/perseverance_wang/article/details/88981055</guid><author>perseverance_wang</author><pubDate>Tue, 02 Apr 2019 20:55:27 +0800</pubDate><description><![CDATA[Leetcode-07 Reverse Integer
题目：
Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321
Example 2:

Input: -123
Output: -321
Example 3:

Input: 120
Output: 21
...]]></description><category></category></item><item><title><![CDATA[2.Leetcode03-longest-substring-without-repeating-characters]]></title><link>https://blog.youkuaiyun.com/perseverance_wang/article/details/88956071</link><guid>https://blog.youkuaiyun.com/perseverance_wang/article/details/88956071</guid><author>perseverance_wang</author><pubDate>Mon, 01 Apr 2019 20:06:19 +0800</pubDate><description><![CDATA[题目来源：Leetcode03-longest-substring-without-repeating-characters
问题描述：求最大无重复字符的子串
实现如下：
class Solution(object):
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype...]]></description><category></category></item><item><title><![CDATA[Leetcode练习01]]></title><link>https://blog.youkuaiyun.com/perseverance_wang/article/details/88932970</link><guid>https://blog.youkuaiyun.com/perseverance_wang/article/details/88932970</guid><author>perseverance_wang</author><pubDate>Sun, 31 Mar 2019 18:22:29 +0800</pubDate><description><![CDATA[Leetcode练习01
题目加粗样式来源于leetcode 01  -------&gt; two-sum
问题描述
给定一个整数数组，和一个确定的数值target，返回数组中的两数之和等于target的下标
例：
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0,...]]></description><category></category></item></channel></rss>