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	<title>
	Comments on: 花花酱 LeetCode 234. Palindrome Linked List	</title>
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		<title>
		By: longspring		</title>
		<link>https://zxi.mytechroad.com/blog/list/leetcode-234-palindrome-linked-list/#comment-11</link>

		<dc:creator><![CDATA[longspring]]></dc:creator>
		<pubDate>Thu, 03 May 2018 06:11:52 +0000</pubDate>
		<guid isPermaLink="false">http://zxi.mytechroad.com/blog/?p=2081#comment-11</guid>

					<description><![CDATA[In reply to &lt;a href=&quot;https://zxi.mytechroad.com/blog/list/leetcode-234-palindrome-linked-list/#comment-10&quot;&gt;longspring&lt;/a&gt;.

第一段comment有个typo，应该说这样slow-&#062;next就会always是指到要reverse的head。]]></description>
			<content:encoded><![CDATA[<p>第一段comment有个typo，应该说这样slow-&gt;next就会always是指到要reverse的head。</p>
]]></content:encoded>
		
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		<item>
		<title>
		By: longspring		</title>
		<link>https://zxi.mytechroad.com/blog/list/leetcode-234-palindrome-linked-list/#comment-10</link>

		<dc:creator><![CDATA[longspring]]></dc:creator>
		<pubDate>Thu, 03 May 2018 06:04:17 +0000</pubDate>
		<guid isPermaLink="false">http://zxi.mytechroad.com/blog/?p=2081#comment-10</guid>

					<description><![CDATA[In reply to &lt;a href=&quot;https://zxi.mytechroad.com/blog/list/leetcode-234-palindrome-linked-list/#comment-9&quot;&gt;longspring&lt;/a&gt;.

如果是这样的话一开始就需要检查 head-&#062;next是不是nullptr
    if (!head &#124;&#124; !head-&#062;next) return true;]]></description>
			<content:encoded><![CDATA[<p>如果是这样的话一开始就需要检查 head-&gt;next是不是nullptr<br />
    if (!head || !head-&gt;next) return true;</p>
]]></content:encoded>
		
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		<item>
		<title>
		By: longspring		</title>
		<link>https://zxi.mytechroad.com/blog/list/leetcode-234-palindrome-linked-list/#comment-9</link>

		<dc:creator><![CDATA[longspring]]></dc:creator>
		<pubDate>Thu, 03 May 2018 05:57:36 +0000</pubDate>
		<guid isPermaLink="false">http://zxi.mytechroad.com/blog/?p=2081#comment-9</guid>

					<description><![CDATA[这题是不是先将fast 初始的时候让他等于 head-&#062;next，这样就不需要检查odd or even了。一快一慢前进的时候，slow指到的就会是要reverse的head。]]></description>
			<content:encoded><![CDATA[<p>这题是不是先将fast 初始的时候让他等于 head-&gt;next，这样就不需要检查odd or even了。一快一慢前进的时候，slow指到的就会是要reverse的head。</p>
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