lc234. 回文链表（反转链表）

给你一个单链表的头节点 head ，请你判断该链表是否为回文链表。如果是，返回 true ；否则，返回 false 。

示例 1：

输入：head = [1,2,2,1]
输出：true
示例 2：

输入：head = [1,2]
输出：false

提示：

链表中节点数目在范围[1, 105] 内
0 <= Node.val <= 9

进阶：你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？

/**
 * 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;
        }
        //奇数,确保slow是后半段
        if (fast != null) slow = slow.next;
        ListNode reverseNode = reverse(slow);
        while (reverseNode != null) {
            if (reverseNode.val != head.val) return false;
            reverseNode = reverseNode.next;
            head = head.next;
        }
        return true;
    }
    private ListNode reverse(ListNode head) {
        ListNode pre = null;
        while (head != null) {
            ListNode next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
}

递归

class Solution {
    ListNode pre;
    public boolean isPalindrome(ListNode head) {
        pre = head;
        return recur(head);
    }
    private boolean recur(ListNode head) {
        if (head == null) return true;
        if (!recur(head.next)) return false;
        if (pre.val != head.val) return false;
        pre = pre.next;
        return true;
    }
}