(easy)LeetCode 234.Palindrome Linked List

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

思想:转置后半段链表节点，然后比较前半段和后半段节点的值是否相等。

代码如下：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isPalindrome(ListNode head) {
        if(head==null || head.next==null) return true;
        
        int len=0,tmp=0;
        ListNode p=head,q=head,r=head,s=head;
        int m1=0,m2=0;
        int n1=0,n2=0;
        while(p!=null){
            len++;
            p=p.next;
        }
        if(len==2){
            if(head.val==head.next.val)
               return true;
            else
               return false;
        }
        p=head;
        if(len%2==1){
           m1=0;
           m2=len/2+1;
           n1=m2-2;
           n2=len-1;
            
        }else{
            m1=0;
            m2=len/2;
            n1=m2-1;
            n2=len-1; 
            
        }
        while(p!=null && tmp!=m2){
                p=p.next;   
                tmp++;
              }  //p到达转置数组的头部
          q=p;
        if(p.next!=null){
             q=p;
             p=p.next;
             q.next=null;
             
             while(p!=null){
                 r=p.next;
                 p.next=q;
                 q=p;
                 p=r;
             }//while 循环后,q指后半段头
          }  
            while(q!=null &&s.val==q.val){
                q=q.next;s=s.next;
            }
            if(q==null)
               return true;
            else
               return false;
           
    }
}

　　运行结果：