题解 | #判断一个链表是否为回文结构#

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 the head
     * @return bool布尔型
     */
    public boolean isPail (ListNode head) {
        // write code here
        ListNode slow = head, fast = head;
        while(fast != null){
            slow = slow.next;
            fast = fast.next;
            if(fast != null)fast = fast.next;
        }
        ListNode rev =  reverse(slow);
        while(rev != null){
            if(head.val != rev.val)return false;
            rev = rev.next;
            head = head.next;
        }
        return true;
    }
    private ListNode reverse(ListNode node){
        ListNode pre = null, cur = node, next;
        while(cur != null){
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}

用快慢指针找到链表的中间节点，翻转后半段链表，再进行比较，时间、空间复杂度最优。