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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?

题解：获得链表一半的位置，再将其反转，最后2个半截的链表比较

public boolean isPalindrome(ListNode head) {		if(head == null || head.next == null)			return true;		ListNode fast=head,slow=head;		while(fast.next!=null && fast.next.next!=null ){ //slow走到一半的位置			slow = slow.next;			fast = fast.next.next;		}		     ListNode font = head;  			slow = reverseList(slow);				while(font!=null){   //两个指针比较						if(font.val != slow.val){				return false;			}			font = font.next;			slow = slow.next;		}		return true;	}	public ListNode reverseList(ListNode head){ //将后半截链表反转				ListNode temp = head.next,pre = head;		while(temp!=null){			ListNode r = temp.next;			temp.next = pre;			pre = temp;			temp = r;		}		head.next = null;		return pre;	}

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