Given the head
of a singly linked list, return true
if it is a palindrome.
Example 1:
Input: head = [1,2,2,1] Output: true
Example 2:
Input: head = [1,2] Output: false
Constraints:
- The number of nodes in the list is in the range
[1, 105]
. 0 <= Node.val <= 9
Follow up: Could you do it in
O(n)
time and O(1)
space?
# 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.
* 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;
}
}
/**
* 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;
}
console.log(slow)
// 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;
};