LeetCode234. 回文链表

思路：【快慢指针 + 反转链表】通过快慢指针找到中间节点 ----> 切成两个链表 ----> 对后半部分进行reverse操作 -----> 依次比较前部分和后部分的值

与LeetCode143. 重排链表解法类似。

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;
    }
}