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：
