LeetCode Reverse Linked List （反置链表）

题意：

　　将单恋表反转。

思路：

　　两种方法：迭代和递归。

　　递归

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode* give_me_your_son( ListNode* far, ListNode* son)
     {
         if(!son)    return far;     //far就是链表尾了
         ListNode* tmp=son->next;
         son->next=far;
         return give_me_your_son( son, tmp);
     }

     ListNode* reverseList(ListNode* head) {
         if(!head||!head->next)   return head;
         return give_me_your_son(,head);
     }
 };

AC代码

python3

　　迭代

 # Definition for singly-linked list.
 # class ListNode(object):
 #     def __init__(self, x):
 #         self.val = x
 #         self.next = None

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

AC代码

　　递归

 # Definition for singly-linked list.
 # class ListNode(object):
 #     def __init__(self, x):
 #         self.val = x
 #         self.next = None

 class Solution(object):
     def reverseList(self, head, pre=None):
         """
         :type head: ListNode
         :rtype: ListNode
         """
         if not head:    return pre
         back=head.next
         head.next=pre
         return self.reverseList(back, head)

AC代码