<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[jackywiki的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/jackywiki</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; jackywiki]]></copyright><item><title><![CDATA[LeetCode 14. 最长公共前缀]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/170037049</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/170037049</guid><author>jackywiki</author><pubDate>Sat, 07 Nov 2026 00:44:16 +0800</pubDate><description><![CDATA[给定字符串数组strs，返回所有字符串的最长公共前缀。如果不存在公共前缀，返回空字符串""。]]></description><category></category></item><item><title><![CDATA[LeetCode 13. 罗马数字转整数]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169950472</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169950472</guid><author>jackywiki</author><pubDate>Wed, 04 Nov 2026 11:30:50 +0800</pubDate><description><![CDATA[给定一个有效的罗马数字字符串s，返回其对应的整数。]]></description><category></category></item><item><title><![CDATA[LeetCode 12. 整数转罗马数字]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169950344</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169950344</guid><author>jackywiki</author><pubDate>Wed, 04 Nov 2026 11:30:21 +0800</pubDate><description><![CDATA[给定整数num，将其转换为罗马数字。按千位、百位、十位、个位依次转换并拼接。XCM最多连续出现三次；VLD不重复使用。]]></description><category></category></item><item><title><![CDATA[LeetCode 11. 盛最多水的容器]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169948153</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169948153</guid><author>jackywiki</author><pubDate>Tue, 03 Nov 2026 17:17:49 +0800</pubDate><description><![CDATA[给定长度为n的数组height，下标i处有一条高度为height[i]的竖线。选择其中两条竖线，与横轴共同组成容器，返回能够容纳的最大水量。注意：容器不能倾斜，水量由两条线中较短的一条决定。]]></description><category></category></item><item><title><![CDATA[LeetCode 10. 正则表达式匹配]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169947779</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169947779</guid><author>jackywiki</author><pubDate>Tue, 03 Nov 2026 17:03:25 +0800</pubDate><description><![CDATA[给定字符串s和模式p返回匹配结果，要求模式匹配整个字符串，而不是其中一部分。a*表示零个或多个a.*才表示零个或多个任意字符。]]></description><category></category></item><item><title><![CDATA[LeetCode 9. 回文数]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169749492</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169749492</guid><author>jackywiki</author><pubDate>Mon, 02 Nov 2026 01:09:44 +0800</pubDate><description><![CDATA[给定整数x，判断其是否为回文数，是则返回true，否则返回false。回文数是指从左向右和从右向左读都相同的整数。：不将整数转换为字符串来解决。]]></description><category></category></item><item><title><![CDATA[LeetCode 8. 字符串转换整数（atoi）]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169666250</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169666250</guid><author>jackywiki</author><pubDate>Mon, 02 Nov 2026 01:07:39 +0800</pubDate><description><![CDATA[实现myAtoi(s)' '0。]]></description><category></category></item><item><title><![CDATA[LeetCode 7. 整数反转]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169656341</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169656341</guid><author>jackywiki</author><pubDate>Sun, 01 Nov 2026 02:14:52 +0800</pubDate><description><![CDATA[给定一个 32 位有符号整数x，将其十进制数字反转后返回，保留正负号。如果反转结果超出，返回0。要求不能使用 64 位整数存储中间结果。]]></description><category></category></item><item><title><![CDATA[LeetCode 6. Z 字形变换]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169170922</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169170922</guid><author>jackywiki</author><pubDate>Sun, 01 Nov 2026 01:42:39 +0800</pubDate><description><![CDATA[将字符串s按指定行数numRows进行 Z 字形排列：先从上往下，再斜向右上，重复这一过程。最后从上到下，逐行读取字符，返回拼接后的字符串。]]></description><category></category></item><item><title><![CDATA[LeetCode 5. 最长回文子串]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169169528</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169169528</guid><author>jackywiki</author><pubDate>Wed, 28 Oct 2026 09:35:07 +0800</pubDate><description><![CDATA[给定字符串s，返回其中最长的回文子串。回文指正着读和反着读相同；子串必须连续。如果存在多个最长答案，返回任意一个即可。]]></description><category></category></item><item><title><![CDATA[LeetCode 4. 寻找两个正序数组的中位数]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169155671</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169155671</guid><author>jackywiki</author><pubDate>Wed, 28 Oct 2026 09:34:26 +0800</pubDate><description><![CDATA[给定两个长度分别为mn的非递减有序数组nums1和nums2，返回它们合并后的中位数。要求算法的时间复杂度为。]]></description><category></category></item><item><title><![CDATA[LeetCode 3. 无重复字符的最长子串]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169075011</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169075011</guid><author>jackywiki</author><pubDate>Tue, 27 Oct 2026 00:06:23 +0800</pubDate><description><![CDATA[给定字符串s，求其中不含重复字符的最长连续子串的长度。注意：子串必须连续，不能跳过中间的字符。]]></description><category></category></item><item><title><![CDATA[LeetCode 2. 两数相加]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/169072254</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/169072254</guid><author>jackywiki</author><pubDate>Mon, 26 Oct 2026 02:03:29 +0800</pubDate><description><![CDATA[给定两个非空链表l1和l2，分别表示两个非负整数。每个节点存储一位数字，数字按照排列，即头节点表示个位。计算两个整数之和，并以相同的逆序链表形式返回结果。除数字0外，输入所表示的整数均不含前导零。]]></description><category></category></item><item><title><![CDATA[LeetCode 1. 两数之和]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/168979626</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/168979626</guid><author>jackywiki</author><pubDate>Sun, 25 Oct 2026 01:45:32 +0800</pubDate><description><![CDATA[给定整数数组nums和目标值target，找出数组中和为target的两个元素，返回它们的下标。每组输入恰好有一个有效答案，同一个位置不能使用两次，答案的下标顺序不限。]]></description><category></category></item><item><title><![CDATA[AcWing 算法基础课笔记【基础算法篇】5.双指针算法与位运算]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/141226582</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/141226582</guid><author>jackywiki</author><pubDate>Sun, 15 Sep 2024 19:35:50 +0800</pubDate><description><![CDATA[先写出暴力解法，在根据单调性优化。公式：n >> k & 1。公式： x & (-x)时间复杂度的算法转变为。]]></description><category></category></item><item><title><![CDATA[leetcode刷题（1）704. 二分查找]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/141269391</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/141269391</guid><author>jackywiki</author><pubDate>Thu, 29 Aug 2024 11:40:50 +0800</pubDate><description><![CDATA[题目链接: 704.二分查找
二分法思路：AcWing 算法基础课笔记【基础算法篇】2.整数二分与浮点数二分
时间复杂度为O(logn)O(logn)O(logn)，空间复杂度O(1)O(1)O(1)
同样使用了二分法，时间复杂度为O(logO(logO(log n)n)n)，空间复杂度O(1)O(1)O(1)
同类题目：
AcWing 789.数的范围]]></description><category></category></item><item><title><![CDATA[AcWing 算法基础课笔记【基础算法篇】4.前缀和与差分]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/141195192</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/141195192</guid><author>jackywiki</author><pubDate>Thu, 15 Aug 2024 16:52:49 +0800</pubDate><description><![CDATA[假设有一个数组AA1​A2​...An​那么可以定义一个前缀和数组SSi​A1​A2...Ai​其中，S0​0。]]></description><category></category></item><item><title><![CDATA[AcWing 算法基础课笔记【基础算法篇】3.高精度运算]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/141182609</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/141182609</guid><author>jackywiki</author><pubDate>Wed, 14 Aug 2024 16:27:10 +0800</pubDate><description><![CDATA[存储时要倒着存，方便进位，如A0对应个位，A1对应十位，以此类推。]]></description><category></category></item><item><title><![CDATA[AcWing 算法基础课笔记【基础算法篇】2.整数二分与浮点数二分]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/141171877</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/141171877</guid><author>jackywiki</author><pubDate>Tue, 13 Aug 2024 23:23:02 +0800</pubDate><description><![CDATA[定义一种性质checkcheckcheck，把数组分为两个部分（两部分没有交点）总结：出现l=midl=midl=mid（求左半部分边界）则向上取整，出现r=midr=midr=mid（求右半部分边界）则向下取整
第二类 求左半部分边界

浮点数二分
思想
与整数二分大致相同，但浮点数二分不需要考虑+1等问题]]></description><category></category></item><item><title><![CDATA[AcWing 算法基础课笔记【基础算法篇】1.快速排序与归并排序]]></title><link>https://blog.youkuaiyun.com/jackywiki/article/details/141019298</link><guid>https://blog.youkuaiyun.com/jackywiki/article/details/141019298</guid><author>jackywiki</author><pubDate>Tue, 13 Aug 2024 18:50:36 +0800</pubDate><description><![CDATA[x对于一个无序数组qnql​qlr/2​qr​。]]></description><category></category></item></channel></rss>