<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[e_dimension的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/e_dimension</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; e_dimension]]></copyright><item><title><![CDATA[力扣LCR讲解]]></title><link>https://blog.youkuaiyun.com/e_dimension/article/details/146294425</link><guid>https://blog.youkuaiyun.com/e_dimension/article/details/146294425</guid><author>e_dimension</author><pubDate>Sun, 16 Mar 2025 13:47:18 +0800</pubDate><description><![CDATA[个人自用题目解析回顾。]]></description><category></category></item><item><title><![CDATA[弗洛伊德算法（求目标间的最短路径）]]></title><link>https://blog.youkuaiyun.com/e_dimension/article/details/146283247</link><guid>https://blog.youkuaiyun.com/e_dimension/article/details/146283247</guid><author>e_dimension</author><pubDate>Sat, 15 Mar 2025 19:56:20 +0800</pubDate><description><![CDATA[个人自用题目解析回顾。]]></description><category></category></item><item><title><![CDATA[力扣LCR讲解]]></title><link>https://blog.youkuaiyun.com/e_dimension/article/details/146190696</link><guid>https://blog.youkuaiyun.com/e_dimension/article/details/146190696</guid><author>e_dimension</author><pubDate>Tue, 11 Mar 2025 22:42:38 +0800</pubDate><description><![CDATA[个人自用题目解析回顾LCR 445.两数相加II。]]></description><category></category></item><item><title><![CDATA[力扣LCR讲解]]></title><link>https://blog.youkuaiyun.com/e_dimension/article/details/146161959</link><guid>https://blog.youkuaiyun.com/e_dimension/article/details/146161959</guid><author>e_dimension</author><pubDate>Mon, 10 Mar 2025 19:58:03 +0800</pubDate><description><![CDATA[1.建立哈希表，利用哈希表的索引高效查询效率，将原数组中的所有数转为相同长度的二进制数，存储到哈希表中，若二进制长度不足最大长度（30），则再左侧补位0即可；如果其出现在哈希表中，那么说明第 k 个二进制位能够取到 1，否则第 k 个二进制位只能为 0。表示 x 从最高位第 30 个二进制位开始，到第 k 个二进制位为止的数，那么。根据以上思路，我们得出代码需要实现的操作有。对我们来说已知，因此我们将所有的。放入哈希表中，随后枚举 i 并计算。个人自用题目解析回顾。]]></description><category></category></item></channel></rss>