Leetcode 234. Palindrome Linked List

Posted 2020-09-29 Deribs4

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

 

Similar Questions 

Palindrome Number Valid Palindrome Reverse Linked List 

Next challenges

Valid Palindrome Reverse Linked List

 

 

思路：3步走。

1.设置fast和slow指针，使得fast遍历列表后，slow位于列表中偏右部。

2.反转slow侧的列表。

3.将fast指向head，依次比较slow侧的列表和fast侧列表每个元素的值。

 

代码：上述1和2的2个操作还是比较经典的。

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public boolean isPalindrome(ListNode head) {
11         ListNode fast = head, slow = head;
12 //        每1次fast都比slow多走1个单位直到fast不能再1次走2步为止
13         while(fast != null && fast.next != null) {
14             slow = slow.next;
15             fast = fast.next.next;
16         }
17         if(fast != null) {//有奇数个node，设置slow侧的node数量少于fast侧的node数量
18             slow = slow.next;
19         }
20         slow = reverse(slow);//slow侧列表反转
21         fast = head;
22         while(slow != null && slow.val == fast.val) {
23             slow = slow.next;
24             fast = fast.next;
25         }
26         return slow == null;
27     }
28     
29     //列表反转
30     public ListNode reverse(ListNode head) {
31         ListNode prev = null;
32         while(head != null) {
33             ListNode next = head.next;
34             head.next = prev;
35             prev = head;
36             head = next;
37         }
38         return prev;
39     }
40 }