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?
来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/palindrome-linked-list
/**
* 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) {
ListNode fast = head, slow = head;
while(fast != null && fast.next != null){
fast = fast.next.next;
slow = slow.next;
}
/* if this list has odd number of nodes */
if(fast != null){
slow = slow.next;
}
slow = reverseList(slow);
fast = head;
while(slow != null){
if(slow.val != fast.val){
return false;
}
slow = slow.next;
fast = fast.next;
}
return true;
}
public ListNode reverseList(ListNode head){
ListNode preNode = null;
ListNode currNode = head;
while(currNode != null){
ListNode nextNode = currNode.next;
currNode.next = preNode;
preNode = currNode;
currNode = nextNode;
}
return preNode;
}
}