给定一个链表的 头节点 head ，请判断其是否为回文链表。
如果一个链表是回文，那么链表节点序列从前往后看和从后往前看是相同的。
示例 1：
输入: head = [1,2,3,3,2,1] 输出: true
示例 2：
输入: head = [1,2] 输出: fasle
提示：
- 链表 L 的长度范围为
[1, 105] 0 <= node.val <= 9
进阶：能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？
注意：本题与主站 234 题相同：https://leetcode.cn/problems/palindrome-linked-list/
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def isPalindrome(self, head: ListNode) -> bool:
if head is None or head.next is None:
return True
slow, fast = head, head.next
while fast and fast.next:
slow, fast = slow.next, fast.next.next
pre, cur = None, slow.next
while cur:
t = cur.next
cur.next = pre
pre, cur = cur, t
while pre:
if pre.val != head.val:
return False
pre, head = pre.next, head.next
return True/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public boolean isPalindrome(ListNode head) {
if (head == null || head.next == null) {
return true;
}
ListNode slow = head;
ListNode fast = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
ListNode cur = slow.next;
slow.next = null;
ListNode pre = null;
while (cur != null) {
ListNode t = cur.next;
cur.next = pre;
pre = cur;
cur = t;
}
while (pre != null) {
if (pre.val != head.val) {
return false;
}
pre = pre.next;
head = head.next;
}
return true;
}
}/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
bool isPalindrome(ListNode* head) {
if (!head || !head->next) return true;
ListNode* slow = head;
ListNode* fast = head->next;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
ListNode* pre = nullptr;
ListNode* cur = slow->next;
while (cur) {
ListNode* t = cur->next;
cur->next = pre;
pre = cur;
cur = t;
}
while (pre) {
if (pre->val != head->val) return false;
pre = pre->next;
head = head->next;
}
return true;
}
};/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func isPalindrome(head *ListNode) bool {
if head == nil || head.Next == nil {
return true
}
slow, fast := head, head.Next
for fast != nil && fast.Next != nil {
slow, fast = slow.Next, fast.Next.Next
}
var pre *ListNode
cur := slow.Next
for cur != nil {
t := cur.Next
cur.Next = pre
pre = cur
cur = t
}
for pre != nil {
if pre.Val != head.Val {
return false
}
pre, head = pre.Next, head.Next
}
return true
}/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function isPalindrome(head: ListNode | null): boolean {
if (head == null || head.next == null) return true;
// 快慢指针定位到中点
let slow: ListNode = head,
fast: ListNode = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
// 翻转链表
let cur: ListNode = slow.next;
slow.next = null;
let prev: ListNode = null;
while (cur != null) {
let t: ListNode = cur.next;
cur.next = prev;
prev = cur;
cur = t;
}
// 判断回文
while (prev != null) {
if (prev.val != head.val) return false;
prev = prev.next;
head = head.next;
}
return true;
}/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {boolean}
*/
var isPalindrome = function (head) {
if (!head || !head.next) {
return true;
}
let slow = head;
let fast = head.next;
while (fast && fast.next) {
slow = slow.next;
fast = fast.next.next;
}
let cur = slow.next;
slow.next = null;
let pre = null;
while (cur) {
let t = cur.next;
cur.next = pre;
pre = cur;
cur = t;
}
while (pre) {
if (pre.val !== head.val) {
return false;
}
pre = pre.next;
head = head.next;
}
return true;
};/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
public class Solution {
public bool IsPalindrome(ListNode head) {
if (head == null || head.next == null)
{
return true;
}
ListNode slow = head;
ListNode fast = head.next;
while (fast != null && fast.next != null)
{
slow = slow.next;
fast = fast.next.next;
}
ListNode cur = slow.next;
slow.next = null;
ListNode pre = null;
while (cur != null)
{
ListNode t = cur.next;
cur.next = pre;
pre = cur;
cur = t;
}
while (pre != null)
{
if (pre.val != head.val)
{
return false;
}
pre = pre.next;
head = head.next;
}
return true;
}
}