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        <title>三叶 - 标签 - zwyyy456&#39;s blog</title>
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        <description>三叶 - 标签 - zwyyy456&#39;s blog</description>
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    <title>3.无重复字符的最长子串</title>
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    <pubDate>Mon, 05 Dec 2022 20:20:31 &#43;0800</pubDate><author>
                    <name>zwyyy456</name>
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    <description><![CDATA[1 问题描述3.无重复字符的最长子串 2 解题思路用arr[96]记录每个字符出现的次数，如果字符次数大于1，说明已经包含了重复字符，则要更新l，]]></description>
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    <title>30.串联所有单词串</title>
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    <pubDate>Mon, 05 Dec 2022 19:55:54 &#43;0800</pubDate><author>
                    <name>zwyyy456</name>
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    <description><![CDATA[1 问题描述30.串联所有单词串 2 解题思路首先，由于words中所有字符串长度相同，要比较words与s: - s从i = 0开始，可以划分为一系列的]]></description>
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    <title>1610.可见点的最大数目</title>
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    <pubDate>Sun, 04 Dec 2022 19:42:50 &#43;0800</pubDate><author>
                    <name>zwyyy456</name>
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    <description><![CDATA[1 问题描述1610.可见点的最大数目 2 解题思路利用atan2函数，即可将斜率转化为$-\pi ~ \pi$的角度; 扩充数组，令angle[n + i]]]></description>
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    <title>303.区域和检索-数组不可变</title>
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    <pubDate>Fri, 02 Dec 2022 16:50:52 &#43;0800</pubDate><author>
                    <name>zwyyy456</name>
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    <description><![CDATA[1 问题描述303.区域和检索-数组不可变 2 解题思路利用前缀和数组，使得求区间和时无需再遍历数组求和。 3 代码 cpp class NumArray { private: vector&lt;int&gt; sum; public: NumArray(vector&lt;int&gt; &amp;nums) { sum.resize(nums.size(), 0); sum[0] = nums[0]; for]]></description>
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    <title>1769.移动所有球到每个盒子所需的最小操作数</title>
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    <pubDate>Fri, 02 Dec 2022 16:27:27 &#43;0800</pubDate><author>
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    <description><![CDATA[1 问题描述1769.移动所有球到每个盒子所需的最小操作数 2 解题思路暴力求解，时间复杂度为$\Theta(n^2)$; 可以考虑利用前缀和来降低]]></description>
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