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此题链接：

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULLOutput: 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

这是一道经典的反转链表的题，有递归和迭代2种做法，其中使用递归可以只使用一个中间临时变量。

Python迭代做法：

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    def reverseList(self, head: ListNode) -> ListNode:        prev = None        curr = head        while curr != None:            nextNode = curr.next            curr.next = prev            prev = curr            curr = nextNode        return prev

python递归做法：

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    def reverseList(self, head: ListNode) -> ListNode:        if(head == None or head.next == None):            return head        node = self.reverseList(head.next)        head.next.next = head        head.next = None        return node

C++迭代做法：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution{
   public:    ListNode *reverseList(ListNode *head)    {
           ListNode *prev = NULL;        ListNode *curr = head;        while (curr != NULL)        {
               ListNode *nextnode = curr->next;            curr->next = prev;            prev = curr;            curr = nextnode;        }        return prev;    }};

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