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(0,head);
    }
};

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

　　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)