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) {
        if (head == null || head.next == null) {
            return true;
        }
        int count = 0;
        ListNode node = head;
        while (node != null) {
            count++;
            node = node.next;
        }

        int haft = count / 2;
        node = head;
        int index = 0;
        ListNode haftNode = null;
        ListNode preNode = head;
        while (node != null) {
            if (index == haft) {
                haftNode = node;
                preNode.next = null;
                break;
            }
            preNode = node;
            index++;
            node = node.next;
        }
        ListNode reverseNode = reverseList(haftNode);
        ListNode currentNode1 = head;
        ListNode currentNode2 = reverseNode;
        while (currentNode1 != null && currentNode2 != null) {
            if (currentNode1.val != currentNode2.val) {
                return false;
            }
            currentNode1 = currentNode1.next;
            currentNode2 = currentNode2.next;
        }
        return true;
    }

    public static ListNode reverseList(ListNode head) {
        if (head != null && head.next != null && head.next.next != null) {
            ListNode preNode = head;
            ListNode current = head.next;
            ListNode nextNode;
            //最开始的头节点指向空
            preNode.next = null;
            while (current != null) {
                //先获取下一个节点
                nextNode = current.next;
                //当前节点指向上一个节点
                current.next = preNode;
                //上一个节点赋值为当前节点
                preNode = current;
                //当前节点往后走一步
                current = nextNode;
            }
            return preNode;
        } else if (head != null && head.next != null) {
            ListNode nextNode = head.next;
            head.next = null;
            nextNode.next = head;
            return nextNode;
        } else {
            return head;
        }
    }
}