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

Example 1:

Input: 1->2Output: false

Example 2:

Input: 1->2->2->1Output: true

Follow up:

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

 

找中点，反转后半部分，再一一比较

time: O(n), space: O(1)

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */class Solution {    public boolean isPalindrome(ListNode head) {        if(head == null || head.next == null) {            return true;        }        ListNode fast = head, slow = head;        while(fast.next != null && fast.next.next != null) {            fast = fast.next.next;            slow = slow.next;        }        ListNode p1 = head, p2 = reverse(slow.next);        slow.next = null;        while(p1 != null && p2 != null) {            if(p1.val != p2.val) {                return false;            }            p1 = p1.next;            p2 = p2.next;        }        return true;    }        private ListNode reverse(ListNode head) {        if(head == null || head.next == null) {            return head;        }        ListNode prev = null, cur = head;        while(cur != null) {            ListNode next = cur.next;            cur.next = prev;            prev = cur;            cur = next;        }        return prev;    }}