【回文链表】快慢指针+反转链表-526互联

题意：判断一个链表是不是回文（中心对称）的

【反转链表】题解1：

得到原链表的反转链表，然后对比原链表与反转链表的内容是否一致即可。

反转链表版本代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode rHead = reverse(head);
        while(head != null) {
            if(head.val != rHead.val) {
                return false;
            }
            head = head.next;
            rHead = rHead.next;
        }
        return true;
    }

    public ListNode reverse(ListNode head) {
        return reverse(head, null);
    }

    public ListNode reverse(ListNode head, ListNode rLastHead) {
        if(head == null) return null;
        ListNode rHead = new ListNode(head.val, rLastHead);
        if(head.next == null) return rHead;
        return reverse(head.next, rHead);
    }
}

【快慢指针+反转链表】题解2：

将链表一分为二，保证第一段子链表节点个数不小于第二段
通过快慢指针得到第二段子链表表头（先得到第一段子链表表尾，然后next得到第二段子链表表头）
快指针：每次移动两步
慢指针：每次移动一步
对第二段子链表进行反转
遍历第二段子链表，看其是否与第一段子链表val相同

快慢指针+反转链表版本代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode fast = head;
        ListNode low = head;
        while(fast.next != null && fast.next.next != null) {
            low = low.next;
            fast = fast.next.next;
        }
        low = low.next;
        ListNode rHead = reverse(low);
        while(rHead != null) {
            if(head.val != rHead.val) return false;
            head = head.next;
            rHead = rHead.next;
        }
        return true;
    }

    public ListNode reverse(ListNode head) {
        return reverse(head, null);
    }

    public ListNode reverse(ListNode head, ListNode rLastHead) {
        if(head == null) return null;
        ListNode rHead = new ListNode(head.val, rLastHead);
        if(head.next == null) return rHead;
        return reverse(head.next, rHead);
    }
}