Leetcode: Palindrome Linked List

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?

先分成大小相同(可能长度差1) 两部分,   reverse一个list. 再比较. 

采用Runner Technique，结束时runner在最后一个element，walker在第一半的list的最后一个element，设置list2的head为walker.next 

这样做，第一半list会比第二半多，或者相等。

技巧来了：因为允许两个list不同的地方在于---第一半List可以多最后那一个element。方便起见，把第二个list reverse，挨个比过去，第一半的最后一个元素可以不比。如果reverse第一个list，要分别讨论:第一组对比不同是因为这个element是多余的那个, 还是因为本身不是palindrome，或者要计算两段的长度，总之很麻烦。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head==null || head.next==null) return true;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode l1 = dummy;
        ListNode l2 = dummy;
        while (l1!=null && l1.next!=null) {
            l1 = l1.next.next;
            l2 = l2.next;
        }
        ListNode head1 = head;
        ListNode head2 = l2.next;
        l2.next = null;

        head2 = reverse(head2); //head1's linkedlist may be longer, so if there length are not the same, it could only be in the end

        while (head1 != null && head2 != null) {
            if (head1.val != head2.val) return false;
            head1 = head1.next;
            head2 = head2.next;
        }
        return true;
    }
    
    public ListNode reverse(ListNode head) {
        if (head == null) return null;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode end = head;
        while (end.next != null) {
            end = end.next;
        }
        while (dummy.next != end) {
            ListNode cur = dummy.next;
            ListNode next = cur.next;
            cur.next = end.next;
            end.next = cur;
            dummy.next = next;
        }
        return dummy.next;
    }
}