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<metaname="description"content="Given a singly linked list, determine if it is a palindrome. Example 1:Input: 1->2Output: falseExample 2:Input: 1->2->2->1Output: trueFollow up:Could you do it in O(n) time and O(1) spac">
<metaname="description"content="题目介绍Given a singly linked list, determine if it is a palindrome.给定一个单向链表，判断是否是回文链表 例一 Example 1:Input: 1->2Output: false 例二 Example 2:Input: 1->2->2->1Output: true 挑战下自己Follow up:Could you">
<metaproperty="og:description"content="Given a singly linked list, determine if it is a palindrome. Example 1:Input: 1->2Output: falseExample 2:Input: 1->2->2->1Output: trueFollow up:Could you do it in O(n) time and O(1) spac">
<metaproperty="og:description"content="题目介绍Given a singly linked list, determine if it is a palindrome.给定一个单向链表，判断是否是回文链表 例一 Example 1:Input: 1->2Output: false 例二 Example 2:Input: 1->2->2->1Output: true 挑战下自己Follow up:Could you">
<h2id="题目介绍"><ahref="#题目介绍"class="headerlink"title="题目介绍"></a>题目介绍</h2><p>Given a singly linked list, determine if it is a palindrome.<br>给定一个单向链表，判断是否是回文链表</p>
<h3id="例一-Example-1"><ahref="#例一-Example-1"class="headerlink"title="例一 Example 1:"></a>例一 Example 1:</h3><p>Input: 1->2<br>Output: false</p>
<h3id="例二-Example-2"><ahref="#例二-Example-2"class="headerlink"title="例二 Example 2:"></a>例二 Example 2:</h3><p>Input: 1->2->2->1<br>Output: true</p>
<h3id="挑战下自己"><ahref="#挑战下自己"class="headerlink"title="挑战下自己"></a>挑战下自己</h3><p>Follow up:<br>Could you do it in O(n) time and O(1) space?</p>
<h2id="简要分析"><ahref="#简要分析"class="headerlink"title="简要分析"></a>简要分析</h2><p>首先这是个单向链表，如果是双向的就可以一个从头到尾，一个从尾到头，显然那样就没啥意思了，然后想过要不找到中点，然后用一个栈，把前一半塞进栈里，但是这种其实也比较麻烦，比如长度是奇偶数，然后如何找到中点，这倒是可以借助于双指针，还是比较麻烦，再想一想，回文链表，就跟最开始的一样，链表只有单向的，我用个栈不就可以逆向了么，先把链表整个塞进栈里，然后在一个个 pop 出来跟链表从头开始比较，全对上了就是回文了</p>
<h2 id="题目介绍"><a href="#题目介绍" class="headerlink" title="题目介绍"></a>题目介绍</h2><p>Given a singly linked list, determine if it is a
<h2id="题目介绍"><ahref="#题目介绍"class="headerlink"title="题目介绍"></a>题目介绍</h2><p>Given a singly linked list, determine if it is a palindrome.<br>给定一个单向链表，判断是否是回文链表</p>
<h3id="例一-Example-1"><ahref="#例一-Example-1"class="headerlink"title="例一 Example 1:"></a>例一 Example 1:</h3><p>Input: 1->2<br>Output: false</p>
<h3id="例二-Example-2"><ahref="#例二-Example-2"class="headerlink"title="例二 Example 2:"></a>例二 Example 2:</h3><p>Input: 1->2->2->1<br>Output: true</p>
<h3id="挑战下自己"><ahref="#挑战下自己"class="headerlink"title="挑战下自己"></a>挑战下自己</h3><p>Follow up:<br>Could you do it in O(n) time and O(1) space?</p>
<h2id="简要分析"><ahref="#简要分析"class="headerlink"title="简要分析"></a>简要分析</h2><p>首先这是个单向链表，如果是双向的就可以一个从头到尾，一个从尾到头，显然那样就没啥意思了，然后想过要不找到中点，然后用一个栈，把前一半塞进栈里，但是这种其实也比较麻烦，比如长度是奇偶数，然后如何找到中点，这倒是可以借助于双指针，还是比较麻烦，再想一想，回文链表，就跟最开始的一样，链表只有单向的，我用个栈不就可以逆向了么，先把链表整个塞进栈里，然后在一个个 pop 出来跟链表从头开始比较，全对上了就是回文了</p>
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