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?

题目：给定一个单链表，判断是否为回文?最好时间复杂度O(n)，空间复杂度为 O(1)。

思路：除去要求，最简单的方法则是用stack存储第一次遍历过的点，再第二次遍历依次取出比较。即可；此时空间复杂度为O(n)；

满足要求的解法：将后半部分链表逆转，指向中间节点。再两端同时比较。最后最好将链表恢复成原始状态；

代码1：利用stack

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        Stack<Integer> stack = new Stack<Integer>();
        ListNode node = head;
        while(node!=null){
            stack.push(node.val);
            node=node.next;
        }
        node = head;
        while(node!=null){
            //int top = stack.pop();
            if(stack.pop()!=node.val)
                return false;
            node=node.next;
        }
        return true;
    }
}

代码2：满足时间复杂度O(n)和空间复杂度要求O(1)

class Solution {
	public boolean isPalindrome(ListNode head) {
		if (head == null || head.next == null)
			return true;
		ListNode node = head;
		//ListNode right = null;
		ListNode cur = head;
		while (cur.next != null && cur.next.next != null) {
			cur = cur.next.next;
			node = node.next;
		}
		cur = node.next;
		node.next = null;
		ListNode temp=null;
		while(cur!=null){
			temp=cur.next;
			cur.next=node;
			node=cur;
			cur=temp;
		}
		temp = node;	//保存右区首节点
		cur = head;
		boolean ret = true;
		while(cur!=null&&node!=null){
			if(cur.val!=node.val){
				ret = false;
				break;
			}
			cur=cur.next;
			node=node.next;
		}
        /*
        
		cur=temp.next;
		temp.next=null;
		node = temp;
		//恢复链表原始状态
		while(cur!=null){
			temp=cur.next;
			cur.next=node;
			node = cur;
			cur = temp;
		}*/

		return ret;
	}
}