反转链表 两种解法（Python）

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双指针 时间复杂度 O(N)

# 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:
        prev = None
        cur = head
        while cur:
            tmp = cur.next
            cur.next = prev
            prev = cur
            cur = tmp
        return prev

递归 时间复杂度O(N)

# 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
        cur = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return cur