LeetCode 206 Reverse Linked List--反转链表--迭代与递归解法--递归使用一个临时变量，迭代使用3个

本文链接：https://blog.youkuaiyun.com/zhangpeterx/article/details/88607026

本文详细介绍了如何使用递归和迭代两种方法反转链表，并提供了Python和C++代码实现。通过实例解析，帮助读者理解反转链表的基本原理和操作步骤。

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Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 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 = None

class 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 = None

class 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;
    }
};