226 lines
6.4 KiB
Markdown
226 lines
6.4 KiB
Markdown
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<p align="center">
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<a href="https://mp.weixin.qq.com/s/RsdcQ9umo09R6cfnwXZlrQ"><img src="https://img.shields.io/badge/PDF下载-代码随想录-blueviolet" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw"><img src="https://img.shields.io/badge/刷题-微信群-green" alt=""></a>
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<a href="https://space.bilibili.com/525438321"><img src="https://img.shields.io/badge/B站-代码随想录-orange" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ"><img src="https://img.shields.io/badge/知识星球-代码随想录-blue" alt=""></a>
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</p>
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<p align="center"><strong>欢迎大家<a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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# 234.回文链表
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[力扣题目链接](https://leetcode-cn.com/problems/palindrome-linked-list/)
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请判断一个链表是否为回文链表。
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示例 1:
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* 输入: 1->2
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* 输出: false
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示例 2:
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* 输入: 1->2->2->1
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* 输出: true
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# 思路
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## 数组模拟
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最直接的想法，就是把链表装成数组，然后再判断是否回文。
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代码也比较简单。如下：
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```CPP
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class Solution {
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public:
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bool isPalindrome(ListNode* head) {
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vector<int> vec;
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ListNode* cur = head;
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while (cur) {
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vec.push_back(cur->val);
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cur = cur->next;
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}
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// 比较数组回文
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for (int i = 0, j = vec.size() - 1; i < j; i++, j--) {
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if (vec[i] != vec[j]) return false;
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}
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return true;
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}
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};
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```
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上面代码可以在优化，就是先求出链表长度，然后给定vector的初始长度，这样避免vector每次添加节点重新开辟空间
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```CPP
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class Solution {
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public:
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bool isPalindrome(ListNode* head) {
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ListNode* cur = head;
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int length = 0;
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while (cur) {
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length++;
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cur = cur->next;
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}
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vector<int> vec(length, 0); // 给定vector的初始长度，这样避免vector每次添加节点重新开辟空间
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cur = head;
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int index = 0;
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while (cur) {
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vec[index++] = cur->val;
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cur = cur->next;
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}
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// 比较数组回文
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for (int i = 0, j = vec.size() - 1; i < j; i++, j--) {
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if (vec[i] != vec[j]) return false;
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}
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return true;
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}
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};
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```
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## 反转后半部分链表
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分为如下几步：
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* 用快慢指针，快指针有两步，慢指针走一步，快指针遇到终止位置时，慢指针就在链表中间位置
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* 同时用pre记录慢指针指向节点的前一个节点，用来分割链表
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* 将链表分为前后均等两部分，如果链表长度是奇数，那么后半部分多一个节点
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* 将后半部分反转 ，得cur2，前半部分为cur1
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* 按照cur1的长度，一次比较cur1和cur2的节点数值
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如图所示：
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<img src='https://code-thinking.cdn.bcebos.com/pics/234.回文链表.png' width=600> </img></div>
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代码如下：
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```CPP
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class Solution {
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public:
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bool isPalindrome(ListNode* head) {
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if (head == nullptr || head->next == nullptr) return true;
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ListNode* slow = head; // 慢指针，找到链表中间分位置，作为分割
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ListNode* fast = head;
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ListNode* pre = head; // 记录慢指针的前一个节点，用来分割链表
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while (fast && fast->next) {
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pre = slow;
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slow = slow->next;
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fast = fast->next->next;
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}
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pre->next = nullptr; // 分割链表
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ListNode* cur1 = head; // 前半部分
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ListNode* cur2 = reverseList(slow); // 反转后半部分，总链表长度如果是奇数，cur2比cur1多一个节点
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// 开始两个链表的比较
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while (cur1) {
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if (cur1->val != cur2->val) return false;
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cur1 = cur1->next;
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cur2 = cur2->next;
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}
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return true;
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}
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// 反转链表
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ListNode* reverseList(ListNode* head) {
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ListNode* temp; // 保存cur的下一个节点
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ListNode* cur = head;
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ListNode* pre = nullptr;
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while(cur) {
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temp = cur->next; // 保存一下 cur的下一个节点，因为接下来要改变cur->next
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cur->next = pre; // 翻转操作
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// 更新pre 和 cur指针
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pre = cur;
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cur = temp;
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}
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return pre;
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}
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};
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```
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# 其他语言版本
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## Java
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```java
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```
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## Python
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```python3
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#数组模拟
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class Solution:
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def isPalindrome(self, head: ListNode) -> bool:
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length = 0
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tmp = head
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while tmp: #求链表长度
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length += 1
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tmp = tmp.next
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result = [0] * length
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tmp = head
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index = 0
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while tmp: #链表元素加入数组
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result[index] = tmp.val
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index += 1
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tmp = tmp.next
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i, j = 0, length - 1
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while i < j: # 判断回文
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if result[i] != result[j]:
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return False
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i += 1
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j -= 1
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return True
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#反转后半部分链表
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class Solution:
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def isPalindrome(self, head: ListNode) -> bool:
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if head == None or head.next == None:
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return True
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slow, fast = head, head
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while fast and fast.next:
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pre = slow
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slow = slow.next
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fast = fast.next.next
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pre.next = None # 分割链表
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cur1 = head # 前半部分
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cur2 = self.reverseList(slow) # 反转后半部分，总链表长度如果是奇数，cur2比cur1多一个节点
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while cur1:
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if cur1.val != cur2.val:
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return False
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cur1 = cur1.next
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cur2 = cur2.next
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return True
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def reverseList(self, head: ListNode) -> ListNode:
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cur = head
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pre = None
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while(cur!=None):
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temp = cur.next # 保存一下cur的下一个节点
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cur.next = pre # 反转
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pre = cur
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cur = temp
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return pre
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```
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## Go
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```go
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```
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## JavaScript
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```js
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```
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-----------------------
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* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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* B站视频：[代码随想录](https://space.bilibili.com/525438321)
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* 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码.jpg width=450> </img></div>
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