<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[AliceRainly的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/AliceRainly</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; AliceRainly]]></copyright><item><title><![CDATA[Linux 基本操作补充：]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/125506083</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/125506083</guid><author>AliceRainly</author><pubDate>Tue, 28 Jun 2022 17:14:20 +0800</pubDate><description><![CDATA[Linux基本操作二]]></description><category></category></item><item><title><![CDATA[Linux 基本命令学习：对文件及文件夹的增删改查：vim编辑器的使用]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/125486989</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/125486989</guid><author>AliceRainly</author><pubDate>Mon, 27 Jun 2022 16:52:44 +0800</pubDate><description><![CDATA[Linux入门，vim编辑器基本操作
初学者笔记]]></description><category></category></item><item><title><![CDATA[课程表Ⅱ——就在课程表Ⅰ的代码中修改了几下...]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/106192309</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/106192309</guid><author>AliceRainly</author><pubDate>Mon, 18 May 2020 14:33:20 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/course-schedule-ii/
那就只解释一下改动的地方吧：
class Solution:
    def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -&gt; List[int]:
        p=[0]*numCourses
        q=[[]for i in range(numCourses)]
        k=[]#建立列]]></description><category></category></item><item><title><![CDATA[课程表——方法一——常规思路，代码简短但是效果不佳]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/106168318</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/106168318</guid><author>AliceRainly</author><pubDate>Sun, 17 May 2020 01:37:46 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/course-schedule/
看官方思路太费劲了，不看了，把思路放在代码里讲
看完题目就感觉，这不就跟玩只狼一样嘛，打完这个boss才能解锁下面的剧情，和boss。（表示只狼真不是人玩的！T_T）

咳嗯！但是想要解锁新剧情，可能需要打多个boss。
所以第一种解法就肯定是按照这个思路来解的，一开始想用递归，后来发现递归太麻烦，不太行。遇到了几个坑，也算是基础不扎实的后果
class Solution:
    def canFinish]]></description><category></category></item><item><title><![CDATA[二叉树——1]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105925789</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105925789</guid><author>AliceRainly</author><pubDate>Mon, 04 May 2020 23:45:18 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/maximum-depth-of-binary-tree/
递归
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
...]]></description><category></category></item><item><title><![CDATA[二叉树--2]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105925734</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105925734</guid><author>AliceRainly</author><pubDate>Mon, 04 May 2020 23:35:31 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/
递归方法
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self...]]></description><category></category></item><item><title><![CDATA[二叉树，梅开二度]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105802500</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105802500</guid><author>AliceRainly</author><pubDate>Mon, 27 Apr 2020 23:18:38 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/symmetric-tree/
把第一题做完其实第二题也差不多了，我们把这个二叉树看成两个，然后把第一题的代码中的right改成left应该就可行.
然而事情总没你想象中的那么简单…

写完代码10分钟，修个bug1小时
# Definition for a binary tree node.
# class TreeNode:
#...]]></description><category></category></item><item><title><![CDATA[二叉树]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105801639</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105801639</guid><author>AliceRainly</author><pubDate>Mon, 27 Apr 2020 22:37:22 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/same-tree/
看完题目就感觉用递归最简单粗暴。搞明白二叉树的构造应该就容易了。
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.l...]]></description><category></category></item><item><title><![CDATA[格雷编码]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105648771</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105648771</guid><author>AliceRainly</author><pubDate>Tue, 21 Apr 2020 00:57:40 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/gray-code/
位运算确实nb，这代码越短想得越多，看网上的方法看的我一愣一愣的。都是神仙
class Solution:
    def grayCode(self, n: int) -&gt; List[int]:
        res = [0]
        for i in range(n):
         ...]]></description><category></category></item><item><title><![CDATA[子集]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105648247</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105648247</guid><author>AliceRainly</author><pubDate>Tue, 21 Apr 2020 00:14:03 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/subsets/
用位运算做的，其他方法迭代递归也可以做，但是这个应该是复杂度较小的一种了
思路嘛，就是转化成二进制，放在代码里解释叭。
后来发现python居然还有bin方法，那代码就还能再短点。
class Solution:
    def subsets(self, nums: List[int]) -&gt; List[L...]]></description><category></category></item><item><title><![CDATA[唯一数字]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105499467</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105499467</guid><author>AliceRainly</author><pubDate>Mon, 13 Apr 2020 22:27:50 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/single-number/
这代码这么少就不用注释了叭，sort是排序
class Solution:
    def singleNumber(self, nums: List[int]) -&gt; int:
        nums.sort()
        i=0
        
        while i&lt...]]></description><category></category></item><item><title><![CDATA[无重复的最长子串]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105498788</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105498788</guid><author>AliceRainly</author><pubDate>Mon, 13 Apr 2020 22:11:22 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/
class Solution:
    def lengthOfLongestSubstring(self, s: str) -&gt; int:
        a=[]
        max_len=0
        k=0#k是...]]></description><category></category></item><item><title><![CDATA[最长公共前缀-python]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105353739</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105353739</guid><author>AliceRainly</author><pubDate>Mon, 06 Apr 2020 22:51:13 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/longest-common-prefix/
class Solution:
    def longestCommonPrefix(self, strs: List[str]) -&gt; str:
        if not strs:
            return ""
        strs.sort()#这应该...]]></description><category></category></item><item><title><![CDATA[循环双端队列]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105205787</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105205787</guid><author>AliceRainly</author><pubDate>Mon, 30 Mar 2020 18:49:02 +0800</pubDate><description><![CDATA[用的是纯列表来做的，效果貌似不太好
class MyCircularDeque:

    def __init__(self, k: int):
        self.l=[]
        self.length=k

        

    def insertFront(self, value: int) -&gt; bool:
        if self.isFull():...]]></description><category></category></item><item><title><![CDATA[滑动窗口最大值]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105199367</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105199367</guid><author>AliceRainly</author><pubDate>Mon, 30 Mar 2020 16:32:42 +0800</pubDate><description><![CDATA[看到力扣上的难度是困难，然后就想着先做出来再说，然后就这样了
class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -&gt; List[int]:
        list= [0] * (len(nums)-k+1)
        if len(nums)==1:
            return ...]]></description><category></category></item><item><title><![CDATA[逆波兰表达式求值]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105054788</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105054788</guid><author>AliceRainly</author><pubDate>Mon, 23 Mar 2020 18:39:41 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/evaluate-reverse-polish-notation/
思路：遍历，取符号，将该符号代入数字元素进行计算，计算完成后删去数字元素，得到的元素取代符号。
需要注意的地方：遍历所设的索引  i  要注意其位置。别的地方都还好，也许是python太强了，感觉比一些简单的题目还容易一点。。。
class Solution:
  ...]]></description><category></category></item><item><title><![CDATA[最小栈]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105047497</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105047497</guid><author>AliceRainly</author><pubDate>Mon, 23 Mar 2020 14:07:36 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/min-stack/
思路：这个比较简单，按照题目要求做就好，都是常规的函数。
class MinStack:

    def __init__(self):
        self.stack=[]
        self.min_stack=[]


    def push(self, x: int) -&gt; Non...]]></description><category></category></item><item><title><![CDATA[有效括号]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/105046725</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/105046725</guid><author>AliceRainly</author><pubDate>Mon, 23 Mar 2020 13:43:09 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/valid-parentheses/
思路：要求是字符串中的括号都成对出现而且不能交叉错杂的情况，通过实例可以看出，有效字符串中只允许出现’()’ ‘[]’   ‘{}’  或者’{[()]}'这两种形式，其余都算无效。即可以遍历这个字符串，将一开始出现的左括号存入一个列表中，之后若出现与其对应的右括号则将该左括号从列表中删除，若最...]]></description><category></category></item><item><title><![CDATA[判断环形链表]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/104901612</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/104901612</guid><author>AliceRainly</author><pubDate>Mon, 16 Mar 2020 17:43:48 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/linked-list-cycle/
最开始想到的处理方式是整个数组，把链表元素一个一个放进去，直到有重复的，
class Solution:
    def hasCycle(self, head: ListNode) -&gt; bool:
        tgt=[]
        while head:
         ...]]></description><category></category></item><item><title><![CDATA[删除链表重复元素]]></title><link>https://blog.youkuaiyun.com/AliceRainly/article/details/104900041</link><guid>https://blog.youkuaiyun.com/AliceRainly/article/details/104900041</guid><author>AliceRainly</author><pubDate>Mon, 16 Mar 2020 15:24:49 +0800</pubDate><description><![CDATA[https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list/
一样，与之前作业一样的思路
class Solution:
    def deleteDuplicates(self, head: ListNode) -&gt; ListNode:
        if head==None:return head
    ...]]></description><category></category></item></channel></rss>