【剑指Offer 24. 反转链表】【力扣刷题】【Python】

方法一：多指针

# 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:
        pre=None
        cur=head
        while cur:
            temp=cur.next
            cur.next=pre
            pre=cur
            cur=temp
        return pre

方法二：递归

# 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 not head or not head.next:
            return head
        else:
             cur=self.reverseList(head.next) 
             head.next.next=head
             head.next=None
             return cur

总结

1. 双指针法相对来说好理解，递归法head.next.next=head这句话不好理解
2. 返回链表时只需要返回head就是返回整个链表