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：

输入：1，2，2，1

输出：true

方法1：
class Solution {
    public boolean isPalindrome(ListNode head) {
        List<Integer> vals = new ArrayList<Integer>();
        // 将链表的值复制到数组中
        ListNode currentNode = head;
        while (currentNode != null) {
            vals.add(currentNode.val);
            currentNode = currentNode.next;
        }
        // 使用双指针判断是否回文
        int front = 0;
        int back = vals.size() - 1;
        while (front < back) {
            if (!vals.get(front).equals(vals.get(back))) {
                return false;
            }
            front++;
            back--;
        }
        return true;
    }
}

时间复杂度为O(n),空间复杂度为O(n).

方法2：
class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null) {
            return true;
        }
        // 找到前半部分链表的尾节点并反转后半部分链表
        ListNode firstHalfEnd = endOfFirstHalf(head);
        ListNode secondHalfStart = reverseList(firstHalfEnd.next);
        // 判断是否回文
        ListNode p1 = head;
        ListNode p2 = secondHalfStart;
        boolean result = true;
        while (result && p2 != null) {
            if (p1.val != p2.val) {
                result = false;
            }
            p1 = p1.next;
            p2 = p2.next;
        }        
        // 还原链表并返回结果
        firstHalfEnd.next = reverseList(secondHalfStart);
        return result;
    }
    private ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode nextTemp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextTemp;
        }
        return prev;
    }
    private ListNode endOfFirstHalf(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }
}

时间复杂度为O(n),空间复杂度为O(1).