leetcode206

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution
    {
        Stack<ListNode> S = new Stack<ListNode>();
        private void SaveList(ListNode node)
        {
            if (node != null)
            {
                S.Push(node);

                if (node.next != null)
                {
                    SaveList(node.next);
                }
            }
        }

        private ListNode GetLastNode(ListNode head)
        {
            while (head.next != null)
            {
                head = head.next;
            }
            return head;
        }

        public ListNode ReverseList(ListNode head)
        {
            SaveList(head);
            ListNode newhead = null;
            while (S.Count > )
            {
                var curnode = S.Pop();
                curnode.next = null;
                if (newhead == null)
                {
                    newhead = curnode;//找到链头
                }
                else
                {
                    var prenode = GetLastNode(newhead);//找到新的链尾
                    prenode.next = curnode;
                }
            }
            return newhead;
        }
    }

https://leetcode.com/problems/reverse-linked-list/#/description

简化的代码：

 public class Solution
     {
         public ListNode ReverseList(ListNode head)
         {
             //ListNode p = head;
             ListNode n = null;
             while (head != null)
             {
                 ListNode temp = head.next;
                 head.next = n;
                 n = head;
                 head = temp;
             }
             return n;
         }
     }

补充一个python的实现：

 class Solution(object):
     def reverseList(self, head):
         """
         :type head: ListNode
         :rtype: ListNode
         """
         temp = ListNode()
         while head != None:
             nextnode = head.next
             head.next = temp.next
             temp.next = head
             head = nextnode
         return temp.next