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?

判断一个链表是不是构成回文。如果是数组时，我们会从中间开始向两边比较是不是相等。那么换作链表，只能单向访问，怎么办？可以从中间node开始，后半部分进行反转，然后从最后一个node开始，跟第一个node依次往中间对比。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) {
return true;
}
//find middle Node
ListNode fastNode = head;
ListNode slowNode = head;
boolean isodd = false;
while (fastNode != null && fastNode.next != null) {
slowNode = slowNode.next;
fastNode = fastNode.next.next;
}
//reverse back part list
ListNode preNode = null;
ListNode currNode = slowNode;
ListNode nextNode = null;
while (currNode != null) {
nextNode = currNode.next;
currNode.next = preNode;
preNode = currNode;
currNode = nextNode;
}
//compare front and back part list
while (preNode != null) {
if (preNode.val != head.val) {
return false;
}
preNode = preNode.next;
head = head.next;
}
return true;
}
}