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 > 0)
            {
                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：

 1 public class Solution
 2     {
 3         public ListNode ReverseList(ListNode head)
 4         {
 5             //ListNode p = head;
 6             ListNode n = null;
 7             while (head != null)
 8             {
 9                 ListNode temp = head.next;
10                 head.next = n;
11                 n = head;
12                 head = temp;
13             }
14             return n;
15         }
16     }

补充一个python的实现：

 1 class Solution(object):
 2     def reverseList(self, head):
 3         """
 4         :type head: ListNode
 5         :rtype: ListNode
 6         """
 7         temp = ListNode(0)
 8         while head != None:
 9             nextnode = head.next
10             head.next = temp.next
11             temp.next = head
12             head = nextnode
13         return temp.next

posted on 2017-04-19 11:25 Sempron2800+ 阅读(148) 评论(0) 编辑收藏举报