题解 | #牛群编号的回文顺序#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @return bool布尔型
     */
    public boolean isPalindrome (ListNode head) {
        // write code here
        if (head == null || head.next == null) return true;
        // 1. 快慢指针找中点
        ListNode slow = head, fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        // 2. 逆序中点(奇数节点时slow中点、偶数节点时slow是上中点)之后的链表
        ListNode revHead = null, cur = slow.next, next;
        slow.next = null;
        while (cur != null) {
            next = cur.next;
            cur.next = revHead;
            revHead = cur;
            cur = next;
        }
        // 3. 头尾比较，判断是否是回文
        ListNode p = revHead;
        cur = head;
        boolean ans = true;
        while (cur != null && p != null) {
            if (cur.val != p.val) {
                ans = false;
                break;
            }
            cur = cur.next;
            p = p.next;
        }
        // 4. 恢复原链表
        ListNode pre = null;
        cur = revHead;
        while (cur != null) {
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        slow.next = pre;
        return ans;
    }
}