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题目描述：
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?
解题思路：
使用O(n)的时间复杂度及O(1)的时间复杂度表明顺序遍历链表以及不能够开辟跟链表相当的空间，这样可以找出中间的位置，然后将后半部分的链表反转，然后跟前半部分的链表逐个位置比对。
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) {
ListNode end = head;
ListNode mid = head;
while(end != null && end.next != null){
end = end.next.next;
mid = mid.next;
}
if(end != null) //in case of odd list
mid = mid.next;
mid = reverseList(mid);
while(mid != null){
if(mid.val != head.val)
return false;
mid = mid.next;
head = head.next;
}
return true;
}
public ListNode reverseList(ListNode head){
ListNode pre = null, next = null;
while(head != null){
next = head.next;
head.next = pre;
pre = head;
head = next;
}
return pre;
}
}
Java [Leetcode 234]Palindrome Linked List
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原文地址：http://www.cnblogs.com/zihaowang/p/5184345.html
