lintcode 容易题：Reverse Linked List 翻转链表

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The head of linked list.
     * @return: The new head of reversed linked list.
     */
    public ListNode reverse(ListNode head) {
        // write your code here
        if( head ==null || head.next ==null)
            return head;
        ListNode  second = head.next;
        head.next = null;
        ListNode res = reverse(second);
        second.next = head;
        return res;
    }
}

"""
Definition of ListNode

class ListNode(object):

    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""
class Solution:
    """
    @param head: The first node of the linked list.
    @return: You should return the head of the reversed linked list. 
                  Reverse it in-place.
    """
    def reverse(self, head):
        # write your code here
        if head == None or head.next == None:
            return head 
        second = head.next;
        head.next = None 
        res = self.reverse(second)
        second.next = head
        return res