思路：【快慢指標 + 反轉連結串列】通過快慢指標找到中間節點 ----> 切成兩個連結串列 ----> 對後半部分進行reverse操作 -----> 依次比較前部分和後部分的值

與LeetCode143. 重排連結串列解法類似。

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 l2 = slow.next;
        slow.next = null;
        l2 = reverse(l2);
        while (head != null && l2 != null) {
            if (head.val != l2.val) {
                
 
return false;
            }
            head = head.next;
            l2 = l2.next;
        }
        return true;
    }
    private ListNode reverse(ListNode head) {
        ListNode cur = head;
        ListNode pre = null, next = null;
        while (cur != null) {
            next = cur.next;
            cur.next 
 
= pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}