<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Floyd 判圈 on 111qqz's blog</title><link>https://111qqz.com/en/tags/floyd-%E5%88%A4%E5%9C%88/</link><description>Recent content in Floyd 判圈 on 111qqz's blog</description><generator>Hugo -- gohugo.io</generator><language>en</language><copyright>© 2015-2026 111qqz</copyright><lastBuildDate>Wed, 05 Apr 2017 07:31:49 +0000</lastBuildDate><atom:link href="https://111qqz.com/en/tags/floyd-%E5%88%A4%E5%9C%88/index.xml" rel="self" type="application/rss+xml"/><item><title>leetcode 287. Find the Duplicate Number (floyd判圈算法找重复元素)</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-04-05-leetcode-287-find-the-duplicate-number/</link><pubDate>Wed, 05 Apr 2017 07:31:49 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-04-05-leetcode-287-find-the-duplicate-number/</guid><description>&lt;p&gt;Given an array &lt;em&gt;nums&lt;/em&gt; containing &lt;em&gt;n&lt;/em&gt; + 1 integers where each integer is between 1 and &lt;em&gt;n&lt;/em&gt; (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.&lt;/p&gt;</description></item><item><title>sgu 455. Sequence analysis （floyd 判圈算法，O(1)空间复杂度求循环节）</title><link>https://111qqz.com/en/post/acm-icpc/2015/2015-07-30-sgu455/</link><pubDate>Thu, 30 Jul 2015 13:19:00 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2015/2015-07-30-sgu455/</guid><description>&lt;h4 class="relative group"&gt;455. Sequence analysis
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&lt;p&gt;比赛的时候逗了，往看空间限制了&amp;hellip;.&lt;/p&gt;</description></item></channel></rss>