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?

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解析：1.链表分为两段，不管是奇数还是偶数；2.第二段反转一下；3.两段对比一下

/**
* 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;

ListNode pre=new ListNode(-1);
pre.next=head;
ListNode r1=pre;

ListNode slow=pre.next,fast=pre.next;
while(slow!=null&&fast!=null&&fast.next!=null){
pre=pre.next;
slow=slow.next;
fast=fast.next.next;
}
pre.next=null;

ListNode head1=r1.next;;
ListNode head2=reverse(slow);

while(head1!=null&&head2!=null){
if(head1.val!=head2.val) return false;
else{
head1=head1.next;
head2=head2.next;
}
}
return true;

}

private ListNode reverse(ListNode head){
if(head==null||head.next==null) return head;

Stack<ListNode> stack=new Stack<>();

ListNode pre=new ListNode(-1);
ListNode res=pre;

ListNode temp=head;
while(temp!=null){
stack.push(temp);
temp=temp.next;
}
while(!stack.isEmpty()){
pre.next=stack.pop();
pre=pre.next;
}

return res.next;
}
}