【leetcode】剑指 Offer II 024. 反转链表（python）

方法一：迭代

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head
        pre = head
        cur = head.next
        pre.next = None
        while cur:
            tmp = cur.next
            cur.next = pre
            pre = cur
            cur = tmp
        return pre

方法二：递归

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head
        p = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return p  

推荐一个递归解法的图文详解【递归解法】