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（8）、链表的回文结构

class Solution {    public boolean isPalindrome(ListNode head) {        if(head==null || head.next==null){            return true;        }    //1、找中间节点        ListNode fast=head;        ListNode slow=head;        while(fast!=null && fast.next!=null){            fast=fast.next.next;            slow=slow.next;        }    //2、反转后半部分        ListNode cur=slow.next;        while(cur!=null){            ListNode curNext=cur.next;            cur.next=slow;            slow=cur;            cur=curNext;        }    //3、slow从后往前走，head从前往后走        while(slow != head){            if(head.val != slow.val){                return false;            }            if(head.next==slow){                return true;            }            head=head.next;            slow=slow.next;        }        return true;    }}

（9、两个链表的第一个公共结点

public class Solution {    public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {        if (headA == null || headB == null) {            return null;        }        ListNode pL = headA;//永远指向长的单链表        ListNode pS = headB;//永远指向短的单链表        int lenA = 0;        int lenB = 0;        //求的lenA  lenB        while (pL != null) {            lenA++;            pL = pL.next;        }        while (pS != null) {            lenB++;            pS = pS.next;        }        pL = headA;        pS = headB;        //差值-》最长的单链表先走len步        int len = lenA - lenB;        if (len < 0) {            pL = headB;            pS = headA;            len = lenB - lenA;        }        //让pL先走len步        while (len > 0) {            pL = pL.next;            len--;        }        //开始一起走  (pL != pS )---->一人一步走        while (pL != pS) {            pS = pS.next;            pL = pL.next;        }        return pL;    }}